N3 ENGINEERING SCIENCE SPOT ON R80 (2)

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA T640(E)(J30)T NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 (15070413) 30 July 2019 (X-Paper) 09:00-12:00 REQUIREMENTS: Properties of water and steam (BOE 173) Calculators may be used This question paper consists of 7 pages, a formula sheet of 2 pages and an information sheet of 2 pages. Please turn over Copmiahireseve’ Downloaded from hlayiso.com

(15070413) -2- DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Answers must be rounded off to THREE decimal places. 5. All the calculations should consist of at least the following THREE steps: e The formula used or the manipulation thereof « Substitution of the given data in the formula « The answer with the correct S! unit 6. All drawings and diagrams must be fully labelled and drawing instruments should be used. v. The constant values, as they appear on the attached information sheet, must be used wherever possible. 8. Keep subsections of questions together. 9. Rule off on completion of each question. 10. Use g = 9,8 m/s? 11. Write neatly and legibly. Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) 3. QUESTION 1: MOTION, ENERGY AND POWER 11 1.2 1.3 14 1.5 Copyright reserved Define the term mass of an object. ow Distinguish between displacement and distance. A Toyota Yaris moves from rest and accelerates for 15 seconds and reaches a velocity of 20 m/s. It continues with the uniform velocity for 15 seconds and the driver brakes and the car stops after 10 s. Calculate: 1.3.1 The acceleration in the first 15 seconds 1.3.2 The total distance of the entire motion 1.3.3 The average velocity in 40 seconds Make a sketch of a velocity-time graph for the total duration of motion of the Toyota Yaris. A flat belt fits around a pulley which has a diameter of 45 cm. The belt has a speed of 17,6 m/s and transmits 9 N per mm belt width. The effective pull in the beit is 423 N and the belt has a width of 18 cm. Calculate: 1.5.1 The power transmitted by the belt in kW 1.5.2 The tight side of the force o* 1.5.3 Pulley speed in r/s Downloaded from hlayiso.com (2) (2) (2) (2) [16] Please turn over

(15070413) adhe QUESTION 2: MOMENTS 2.1 State TWO conditions for a beam to be in equilibrium. oe (2) 2.2 In FIGURE 1 a light horizontal beam ABCDE of a uniform cross section is loaded as shown: 5 kN 10 kN 8 kN 5 kN | Z || \ A | + om 4m fim ram | f \ r T 1 B | Cc D E FIGURE 1 2.2.1 Calculate the reactions of the supports and motivate your answer. (6) 2.2.2 Draw a shear force diagram using a suitable scale. Show all the main values on the diagram. . (4) [12] Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -5- QUESTION 3: FORCES 3.1 3.2 Define the term triangle of forces. x o* The forces shown in FIGURE 2 are in equilibrium and in the same plane. 100 N 130° 60 N Lo 80N ¥ 70N FIGURE 2 Calculate: 3.2.1 The sum of the horizontal components in magnitude and direction 3.2.2 The sum of the vertical components in magnitude and direction e* 3.2.3 The magnitude and direction of the equilibrant force QUESTION 4: FRICTION 44 4.2 43 Copwightreseve? Downloaded from hlayiso.com oF Name FOUR principles of kinetic friction. An object is pushed upwards against a slope by a pushing force (P) which forms an angle of 15° with the slope. The object weighs 500 N, and the slope forms an angle of 20° with the horizontal. The coefficient of friction between the surfaces in contact is 0,2. Calculate: 4.2.1 The mass of the object 4.2.2 The smallest force P needed to push the object upwards Give TWO applications of the angle of repose. (3) (3) (6) [14] (1) (6) (2) [13} Please turn over

(15070413) 6- QUESTION 5: HEAT 5.1 5.2 5.3 State TWO applications of steam. A rectangular brass sheet has the following dimensions at 10 °C as indicated on the diagram in FIGURE 3. The coefficient of linear expansion is 20 x 10°/°C. The sheet temperature is then raised to 120 °C. 450 mm Brass sheet 350 mm FIGURE 3 Determine the following: 5.2.1 The area coefficient of expansion 5.2.2 The increase in area in m? at 120 °C 5.2.3 The dimensions at 120 °C o* Calculate the volume of water in litres, required to cool 45 steel shafts. The initial temperature of the shafts is 800 °C and that of water is 35 °C. The mass of each steel shaft is 300 g. The final temperature of the water and steel shafts is 100 °C. & QUESTION 6: HYDRAULICS 6.1 6.2 Copyright reserved The following information refers to a single-acting hydraulic press: Diameter of the ram = 5,5cm Force applied to the plunger = 0,4 kN Diameter of the plunger = 1,2cm Plunger stroke = 0.2m Calculate: 6.1.1 The force exerted by the ram 6.1.2 The distance in mm that the ram piston will move after 20 pumping strokes 6.1.3 The volume of the liquid displaced after 20 pumping strokes of the plunger & Name THREE elementary experiments dealing with pressure in liquids. Downloaded from hlayiso.com (3) (5) (4) [15] (3) (3) (3) Please turn over

(15070413) 7. 6.3 State whether the following statement is True or False: se The symbol for the SI unit of pressure is N/m?. QUESTION 7: ELECTRICITY 7A 7.2 7.3 State TWO very important uses of electroplating in the manufacturing industry and in engineering. A circuit consists of a 6 V battery and two parallel resistors of 4 Q and6 9 and a resistor of 0,6 Q in series. Ignore the internal resistance. Calculate: & 7.2.1 The total resistance of the circuit 7.2.2 The current in the circuit 7.2.3 Current through the 4 resistor A single transformer has a supply voltage of 220 V and a primary current of 20 A at full load. There are 50 primary turns and 280 secondary turns on the transformer. Caiculate the following: 7.3.1 The secondary voltage & 7.3.2 The voltage per turn QUESTION 8: CHEMISTRY 8.1 8.2 8.3 Copyright reserved State TWO ways in which atoms of different elements differ from one another. Name TWO types of corrosion. Name the component elements of the following: 8.3.1 Brass o* 8.3.2 Limestone (2 * 1) TOTAL: Downloaded from hlayiso.com (1) [12] (2) (2) (3) (2) (1) [12] (2) (2) [6] 100

(15070413) io ENGINEERING SCIENCE N3 FORMULA SHEET All the formulae needed are not necessarily included. Any applicable formula may also be used. We Fis Wty. ty EM iy =I, Vp EMV 7-0 EMy U2 =i Vp TMy.V2 W=p.V D, =(D+¥ wv Pe 7 Iryairvce = hg + XM yg a P=la«aTn..T=F yr - Uitset/Output 100% ; InsetInput p= Frans _ Fp, ae aD F=ma Aram — APL 4 F, Vraa =Vpp xn we Np Aran» Aras = App: Eee f=tan® Fy = Feos@ Np = Fo = Frsina ..a=0 Fy = Fsin@ Fy = wsin® LF y = Fycos@, +... + F,cos0, Fo = weos6 ZF, = FsinO; +... + F,sin0, Freosa. = F, + Fy ...a=0 R= SF + 3FF F.=T, -T; tang = 22Y Tr SFy =! = tension ratio T, Q=m.c.At..tp=toLAt P=F.y mww = Q= miwv an N P= 2 ver.d.n..n=— t 60 AL=Lo.a. At... ly=Lo+ AL W, =F, 8 AA=Ay. fp. At... Ap = Ay + AA AE, =m.g.Ah 4 1 Dasev -u AEx = =.m. Av? 2 i s=ut+—at 2 2 Q=I Rut veutat m=Il.z.t EtFsSdF Vp Np _ Is Vs Ng Ip M=F.1s Copyright reserved Please tum over Downloaded from hlayiso.com

(15070413) -2~ X CIVM = SACHWM Pass = Paras + Pyer P=Oxgexh i i i a Rear Ry R, Reap = Rp + Ry V,-V,=—-e(U;-U>) V=IxR Copwiantresese? Downloaded from hlayiso.com

(15070443) INFORMATION SHEET PHYSICAL CONSTANTS CONSTANTS QUANTITY KONSTANTE HOEVEELHEID Atmospheric pressure 101,3 kPa Atmosferiese druk Density of copper 8 900 kg/m* Digtheid van koper Density of aluminium 2770 kg/m Digtheid van aluminium Density of gold 19 000 kg/m Digtheid van goud Density of alcohol (ethy!) 790 kg/m? Digtheid van alkohol (etiel) Density of mercury 13 600 kg/m? Digtheid van kwik Density of platinum 21 500 kg/m? Digtheid van platina Density of water 1 000 kg/m* Digtheid van water Density of mineral oil 920 kg/m* Digtheid van minerale olie Density of air 1,05 kg/m* Digtheid van lug Electrochemical equivalent of silver 1,118 mg/C Elektrocherniese ekwivalent van silwer Electrochemical equivalent of copper 0,329 mg/C Elektrochemiese ekwivalent van koper Gravitational acceleration 9,8 m/s Swaartekragversnelling Heat value of coal 30 MJ/kg Warmtewaarde van steenkool Heat value of anthracite 35 Mi/kg Warmtewaarde van antrasiet Heat value of petrol 45 Mi/ke Warmtewaarde van petrol Heat value of hydrogen 140 Mi/kg Warmtewaarde van waterstof Linear coefficient of expansion of copper | 17 x 10°/°C Line€re uitsettingskoéffisiént van koper Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -2- Linear coefficient of expansion of 23 « 10°C Lineére uitsettingskoéffisiént van aluminium aluminium Linear coefficient of expansion of steel | 12 « 10°/°C | Lineére uitsettingskoéffisiént van staal Linear coefficient of expansion of lead [54% 10°/°C | Lineére uitsettingskoéffisiént van lood Specific heat capacity of steam 2 100 I/kg.°C Spesifieke warmtekapasiteit van stoom Specific heat capacity of water 4 187 J/kg.°C Spesifieke warmtekapasiteit van water Specific heat capacity of aluminium 900 Wkg.°C Spesifieke warmtekapasiteit van aluminium Specific heat capacity of oil 2 000 I/kg.°C Spesifieke warmtekapasiteit van olie Specific heat capacity of steel 500 Jkg.°C Spesifieke warmtekapasiteit van staal Specific heat capacity of copper 390 Ikg.°C Spesificke warmtekapasiteit van koper Copyright reserved Downloaded from hlayiso.com

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digher education & traini ing Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA DELINE | AMENDED MARKING GUI NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 30 JULY 2019 This marking guideline consists of 8 pages. Chief Marker: Lb he OF SORBAE. Stonature, L-Z Chief Marker, WW Aele some NE, chiet Maker: LT Bro ugh Signature: rough Internal Moderator: / Srempwé Signature: Yoav DMCM: 6 : ? OW, fUtH LECCE signature: & (ING | ? MCM: M NTA MS 0 Xt Signature: H { ? Copyright reserved Please turn over Downloaded from hlayiso.com

MARKIPIG GUIDEL WWE -o ENGINEERING SCIENCE N2 1.5.3 v= nDN 47,6 = 1 (0,45)x NY N = 12,46 sw (2) 116] QUESTION 2 2.1 e The sum of the upward forces is equal to the sum of the downward forces. e The sum of the anticlockwise moments is equal to the sum of the clockwise moments. ® The vector sum of forces on the body must be zero (Any two) (2) 22 2.2.1 Moment about B »; ACM = >»; cM 5x14+Cx4=10%24+8x54+5x 6% 4¢=90-5 C= 21,25 kNY Moment about C Yiacm = X CM 5x54+10x2=Bx4t+Bxuit+5x2V 4B = 45-18 B= 6,75 kNv » Fup = > Fdown¥ 21,25+6,75=5+104+8+5 28 KN = 28 kNv (6) 2.2.2 1,75 kN 13 kKNY SKN -5 kKNY ~8,25 kN (4) OR Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE ENGHVEEF 22 2.2.4 Moment about B >acm = Som Sx14Cx%4=40x24+8x545x6¥ 4€=150-—5 C = 36,25 kNVv Moment about G DAcM =) cM 5x54+40x2=8x44+8x14+5x2V 4B = 105-18 B = 21,75 kNv Fup =)" Fdownv 36,25 + 21,75=5+40+8+5 58 KN = 58 kN“ (6) 2.2.2 +16,75 kKN\ ~% 413 KNY +5kN -5 kNY -23,25 kNY (4) [12] QUESTION 3 3.1 Hf three forces acting on a point are in equilibrium,“ they can be represented in magnitude and direction by the sides of a triangle.” (2) 320 3.2.41 > Hz = 100cos50° + 80cos40° — 60 = 65,562 Nv, East’ (3) 322 'S"¥, = 100sin50* — 80sin4o" ~ 707 =-44,819 N,~ South” @) Copyright reserved Please turn over Downloaded from hlayiso.com

MARRING GUIDELINE & NG = 65,562? + 44,8192 « = tan™1(44,819/65,562) = TOAMITNY E 34,36" S¥ = 34,36° v E=79417N~ W3436°NY (6) {14} QUESTION 4 Dependent on the nature of the surfaces in contact Independent of the slip speed Independent of the size of the areain contact v Proportional te the perpendicular force between the surfaces “ Opposite direction which the body is moving (Any four) (4) 41 eo 6@ 3 6 42 4.2.1 F=mg 500 98 m = 51,02 kgv (1/2) () v (1/2) 4.2.2 500cos20° + PsiniS°v 515° — 500sin20°~ Fy E, = fl Po A Fy 02 Px 0,966 — 171,01 469,846 + Px 0,259 93,969 + 0,052P = 0,966P ~ 171,01Y 0.9142P = 263,979 P = 288,574 Nv (Answer can deviate + 2%) 9) Copyright reserved Please turn over Downloaded from hlayiso.com

MAKING GUILKELHSE, 2 SCIENCE N3 QUESTICN 4 4.4 itis the quantity of matter that a body possesses. “ (1) 1.2 Displacement is a vector quantity (magnitude and direction) Vv Distance is a scalar quantity. (magnitude only) ¥ (1) 13 1.3.1 veutat 20 = 0+ ax1bv a= 1,333 ms? (2) 1.3.2 Distance = AA+ eA +AA =0,5x 15x 20+ 15x 20+0,5x 10x 20” = 550m “ (2) 1.3.3 _ As ~ At _550m 405 =13,76m/s ¥ (2) 14 20 v (m/s) 0 15 30 40 v t(s) (3) 15 1.6.4 P=Exv = 423 N x 17,6 m/s” = 7 444,8 WV (1/2) = 7,445 kw ¥ (1/2) (2) 15.2 T, =9Nx180mm = 1620 NV (1) Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELIRE +f ENGINEERING SCIENCE NS 43 « Vehicles are tested against a specified angle of repose to ascertain “ whether they can come to a stop and then start again ~ e In the design of chutes in silos to convey grain « In the design of scoops for power shovels e In the design of railway carriages that transport iron ore e Conveyor belt transporting the material up the incline (Any 2 = 1) (2) [13] QUESTION § 51 Acts as a power source for locomotives, ships and hoists “ Generates electricity turbines “ Used to heat buildings Cook canned food in factories Medical equipment, cans and bottles are sterilised with steam (Any 2 x 1) (2) 5.2 5.2.1 B =2a =2x20x10-° = 40 x 1076/°CY or 0,0000 4 /°C (1) 5.2.2 Ag = 0,45 0,35 = 01575 m?v AA =2xAqgxaxdt = 0,1 575 x 40 x 1078 (120 °C — 10°C)” = 6,93 x 107* m?v or 0,000 693 m? (3) §.2.3 Al= l,xaxAt = 0,45 x 20 x 10° x 110” =9,9x10-*m V or 0,000 990m i, =0,45+9,9 x 107+ = 0,45099 mv Ab= b,xaxAt = 0,35 x 20x 10*x 110 =7,7x10-V 0,000 770m + by = 0,35 + 7,7 x 1074 = 0,35077 mv (8) 5.3 AH gained = Altose mx 4.187(100 °C — 35 °C)“ = 13,5 x 500 x (800 °C - 100 °C)” m= 17,361 kgv v= 17,3611“ (4) [15] Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE oe ENGINEERING SCIENCE N3 CUIESTION 6 6.4 6.2 6.3 6.1.4 Ff De F = 400 x 0,0557/0,012?” F = 8 402,778 NV 6.1.2 exh =d’xhxn 5,57 x H = 1,2? x 20x 20V H = 19,041 cmv H = 190,413 mmv 613 nd? Va xhen 2 okey 0,2 x 20” = 4,524 x 107m?’ or 0,000 4524 m>. v = 0,452 lv Experiment to show that pressure in a liquid is directly proportional to depth. v Experiment to show upward and downward pressure at a point in a liquid. Experiment on the relationship between directional pressure in a liquid. “ Experiment to show that pressure in a liquid is independent of the shape and size of the container “ (Any 3 x 1) True QUESTION 7 7.1 7.2 Copyright reserved Coats low-cost metals with a thin layer of precious metais. “ Chromes tips, hub caps and lettering on cars “ Protects metals against rust, blemish and wear Plates the contact points of electric conductors with metals Builds up worn components by plating metals onto them {Any 2 x 1) 724 Rp =Rpy Rs = [1/4 + 1/67'+0,60V 249+06 30” Downloaded from hlayiso.com (2) (3) (3) (3) (1) [12] (2) (2) Please turn over

MARKING G 722 Vv a Ip IDELINE +8 ENGINEERING SCIENCE N3 =2 Av 7.2.3 Voaratiet = 1px Rp (2) @) 2) (1) [12] (Any 2 * 41) (2) (Any2*1) (2) (2x4) @) 16} TOTAL: 109 Please turn over =2Ax2,40 =4,8Vv v hee = t8ty an = 1,2 Av 7.3 7.3.41 Ny _ N,V, 50 _ 220, 280 «~V; Ve = 1232 VV 7.3.2 Voltage per turn = a OR =4,4v QUESTION 8 8.1 e Structure of their core. (Different numbers of protons and neutrons in their nucleus) ” e The number of electrons outside the core v e The arrangement of the outermost electrons. @ Atomic number 8.2 e Surface corrosion ¥ e Pure corrosion ¥ e Intergranular corrosion e Pure electrochemical corrosion. e Chemical corrosion e Pitting corrosion and Stress corrosion 8.3 8.3.1 Cu and Zn orCopperandZinc “ 8.3.2 Ca, Cand O or Calcium, Carbon and Oxygen Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA T640(E)(A1)T NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 (15070413) 1 April 2019 (X-Paper) 09:00-—12:00 REQUIREMENTS: Properties of water and steam (BOE 173) Calculators may be used. This question paper consists of 8 pages, a formula sheet of 2 pages and an information sheet of 2 pages. Copyright reserved Please turn over Downloaded from hlayiso.co

(15070413) -2- TS40(E) AIT DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4, Answers must be rounded off to THREE decimal places. 5. All the calculations should consist of at least the following THREE steps: 5.1. The formula used or the manipulation thereof 5.2 Substitution of the given data in the formula 5.3. The answer with the correct SI unit 6. All drawings/diagrams must be fully labelled and drawing instruments must be used. 7. The constant values, as they appear on the attached information sheet, must be used wherever possible. 8. Keep subsections of questions together. 9. Rule off on completion of each question. 10. Use g = 9,8 m/s? 14. Write neatly and legibly. Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -3- TE40(EKA1 YT QUESTION 1: MOTION, ENERGY AND POWER 1.1 Distinguish between elastic collision and inelastic collision. (2) 1.2 Two completely elastic balls of 2 kg and 5 kg respectively move in the same Straight line but in opposite directions and collide. Before the collision the 2 kg ball moves at 4 m/s to the right and the 5 kg ball moves at 7 m/s to the left. After the collision the 2 kg ball moves at 3 m/s to the right. os La Calculate the velocity and direction of the 5 kg ball after collision. (3) 1.3 In FIGURE 1 a graph is shown that represents the motion of a train at rest. 20 V (m/s) 10 ! t t I I I I 1 I t t i] 7 0 30 60 t(s) FIGURE 1 Determine: 1.3.1 The acceleration of the train during the first 30 seconds. (2) 1.3.2 The total distance the train has travelled after 60 seconds. (3) 1.4 The driver pulley of a diesel motor has a diameter of 6 cm and rotates at 300 r/min. The flat belt has a thickness of 5 mm. The tension in the tight side is 1 400 N and the tension ratio is 3:1. Consider the belt thickness and calculate the following: 1.4.1 The tension in the slack side 1.4.2 The belt velocity in m/s 1.4.3 The power transmitted in KW (3 * 2) (6) [16] Piease turn over Downloaded from hlayiso.com Copyright reserved

(15070413) “4. T640(E\(A1)T QUESTION 2: MOMENTS 2.1 Define the moment of a force. (2) 2.2 FIGURE 2 shows the shear force diagram of a light horizontal beam ABCDE with a uniform cross section. All the values of the loads are in KN. eal \ \ | | 2m | 3m | 3m 14m i | LL ; + T i dit | | I ' I ; \ I ! \ ; ! | | 30 kN 24,313 KN ! A B 14,313 KN c D EOkN + 5,687 kN - 19,687 kN -34,687 KN 2.2.1 2.2.2 2.2.3 Copyright reserved FIGURE 2 At which TWO positions is the beam supported and what is the magnitude of each support? (2) Calculate the magnitude of the distributed loads between AB and CD. (4) Draw the light horizontal beam ABCDE indicating ALL value loads and supports at the CORRECT positions. a (4) [12] Please turn over Downloaded from hlayiso.com

(45070413) -5- TO40(EMA1)T QUESTION 3: FORCES 3.1 Distinguish between the equilibrant and the resultant force. (2) 3.2 Two ropes support a mass with a weight of 500 N as shown in FIGURE 3. TTT FIGURE 3 = Calculate the tractive forces X and Y in the ropes. (6) Name TWO conditions for STATIC equilibrium if the forces acting on the framework or object are in equilibrium. (2) 3.4 Determine the magnitude and nature of the forces in members AB and BC in the structure shown in FIGURE 4. if KN 10. KN A 30° i c 13,75 kl 21,25 kN (4) FIGURE 4 [14] Copyright reserved Downloaded from hlayiso.com Piease turn over

(15070413) -6- T640(EA1)T QUESTION 4: FRICTION 4.1 State THREE advantages of friction as applicable to engineering. (3) 4.2 A body with a mass of 80 kg is sliding at a constant velocity down an incline without any external force acting on it. The angle of inclination is 30°. ta Calculate the following: 4.2.1 The weight component perpendicular to the sliding plane (2) 4.2.2 The weight component parallel to the sliding plane (2) 4.2.3 The friction force (1) 4.2.4 The coefficient of friction of the plane (2) 4.2.5 The external force required to pull the body up the plane at constant velocity (2) [12] QUESTION 5: HEAT 5.1 Explain what is meant by the specific heat capacity of a substance. (1) 5.2 The following data refer to a copper rod , 4 m in length, being heated through burning coal: Initial temperature of the rod = 20°C Amount of coal burned =750g Percentage of heat transferred to the rod = 30% Mass of the rod =6 kg Calculate the following : §.2.1 The heat energy transferred to the rod (2) §.2.2 The final temperature of the rod (2) 5.2.3 The final length of the rod (3) 5.3 Describe the effect of the change in pressure on the saturation temperature of a substance. (1) 5.4 The total enthalpy of 1 kg of wet steam at a pressure of 900 kPa is 2 600 kJ. Calculate the dryness factor of the steam. (2) Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -T- TS40(EQA1)T 5.5 A mild steel bar with dimensions of 200 cm x 120 mm is heated from 15 ‘C to 70 ‘C and the area increases to 0,2430314 m*. rd Calculate from this the coefficient of linear expansion of steel. (4) [15] QUESTION 6: HYDRAULICS 6.1 Name TWO types of pressure readings which can be taken on a liquid. (2) 6.2 A hydraulic press has a plunger diameter of 25 mm. The force acting on the plunger is 70 N while the ram has a diameter of 12 cm. Caiculate the following: 6.2.1 Mass in kg that can be raised (4) 6.2.2 Pressure in the liquid (3) 6.2.3 Distance that the ram piston will rise after 20 strokes if the plunger stroke is 80 mm (3) [12] QUESTION 7: ELECTRICITY 7 What is meant by the EMF of a cell? (1) 7.2 A battery has an internal resistance of 0,8 Q and is connected in series with a 4 Q resistor and an unknown resistor. When the circuit is closed, the potential difference across the 4 © resistor is 8 V and that across the unknown resistor is 10 V. Calculate the following: 7.2.1 Current through the resistors 7.2.2 Resistance of the unknown resistor (2 x 2) (4) Please turn over Copwigntreseved Downloaded from hlayiso.com

(15070413) -8- TE40(EKA1)T 7.3 74 A current of 8 A flows through a silver nitrate solution for 2 hours in order to electroplate an object. The electrochemical equivalent for silver is 0,001118 g/C. Calculate the following: 7.3.1 Amount of electric charge required 7.3.2 Mass of silver deposited (2 x 2) A single-phase transformer has a supply voltage of 220 V and a primary current of 3 A at full load. The secondary current is 0,2 A and there are 300 turns on the primary coil. Calculate the following: 7.4.1 Turns ratio 74.2 Number of turns in the secondary (2x2) QUESTION 8: CHEMISTRY 8.1 What is an ion? 8.2 State THREE different types of corrosion. 8.3 State TWO properties of brass. TOTAL: Copyright reserved Downloaded from hlayiso.com (4) (4) [13] (1) (3) (2) [8] 100

(15070413) ENGINEERING SCIENCE N3 FORMULA SHEET -1- TO40(E)A1)T All the formulae needed are not necessarily included. Any applicable formula may also be used. We F.s W=p.V P=— t _ Vitset/Output Inset/Input 00% F=ma a Nez fos tan® Np =Fo t Fysina...a=0 Fy = wsin@ Fo = woos? Freosa = F, + Fy ...a=0 F.=T,~-T, q, ; . — = fension ratio 2 P=F,.¥ N ver.d.n..n=— W, =F, .8 4E, =m.g.4h i 2 AER = ze Mm) .ty Emmy My =p Vj bg .V> D,=(D+t Maarivet = hy + XA gy P=2atlinwwTaF ar 2 patra Fre 4 BPO Aram — Api, 4 Vrgsy = Vppxn Aran» rsa = Apy. Lpy Fy = Feos@ Fy = Fsind Fy = Fycos@; +... + F,cos6, LF, = Fysin6) +... + F,sin®, R= JSF? + SF Fy fang = Ze ¥ SF y Q=m.c. Mute =lto£Mt mww = O= miv pik t AL=Lo.a. At... Ly=Lo+ AL AA= Ay. fp. At... Ap = Ag+ AA “pmiantreseved Downloaded from hlayiso.com’*""""

(18070413) O=PF Rut m=l.zt Vp Ne ots Vs Ns Ip 2 CWM = SACWM Paps = Pary + Puer A P=6bxgexh i I i =—+.0¢e0e"° Roar R, R, Copyright reserved Downloaded from hlayiso.com i > SEU +S 2 veutar S*F=X4F M=F.is TE40(EKA1)T

(15070413) INFORMATION SHEET TE40(EWA1\T PHYSICAL CONSTANTS CONSTANTS/ QUANTITY KONSTANTE HOEVEELHEID Atmospheric pressure 101,3 kPa | Atmosferiese druk Density of copper 8 900 kg/m? | Digtheid van koper Density of aluminium 2770 kg/m? | Digtheid van aluminium Density of gold 19 000 kg/m?® Digtheid van goud Density of alcohol (ethyl) 790 kg/m? Digtheid van alkohol (etiel) Density of mercury 13 600 kg/m® Digtheid van kwik Density of platinum 21 500 kg/m* | Digtheid van platina Density of water 1 000 kg/m* | Digtheid van water Density of mineral oil 920 kg/m? | Digtheid van minerale olie Density of air 1,05 kg/m? | Digtheid van lug Electrochemical equivalent of silver 1,118 mg/C | Elektrochemiese ekwivalent van silwer Electrochemical equivalent of copper 0,329 mg/C | Elektrochemiese ekwivalent van koper Gravitational acceleration 9,8 m/s* | Swaartekragversnelling Heat value of coal 30 Mu/kg | Warmtewaarde van steenkool Heat value of anthracite 35 MJ/kg | Warmtewaarde van antrasiet Heat value of petrol 45 MJ/kg | Warmtewaarde van petrol Heat value of hydrogen 140 MJ/kg | Warmtewaarde van waterstof Linear coefficient of expansion of S)0 oe ee oe copper 17 « 10°7°C | LineGre uitsettingskoéffisiént van koper Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -2- T640(EA1)T Linear coefficient of expansion of 5 Lineére uitsettingskoéffisiént van . 23 * 10°C , aluminium aluminium Linear coefficient of expansion of 5 oo. sees su 12 x 10°/°C | Lineére uitsettingskoéffisiént van staal steel Linear coefficient of expansion of 5 oo. . cee sae 54 x 10°/°C | Lineére uitsettingskoéffisiént van lood lead Specific heat capacity of steam 2 100 J/kg.°C Spesifieke warmtekapasiteit van stoom Specific heat capacity of water 4 187 J/kg.°C Spesifieke warmtekapasiteit van water Specific heat capacity of aluminium 900 J/kg.°C Spesifieke warmtekapasiteit van aluminium Specific heat capacity of oil 2.000 d/kg.°C Spesifieke warmtekapasiteit van olie Specific heat capacity of steel 500 dikg.°C Spesifieke warmtekapasiteit van staal Specific heat capacity of copper 300 Jikg.°C Spesifieke warmtekapasiteit van koper Copyright reserved Downloaded from hlayiso.com

& training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 1 APRIL 2019 This marking guideline consists of 9 pages. a \ fp) i Chief Marker: p a) proud Signature: _/, Doug) NS x. bey | Chief Marker: (v\s MA Baveve Signature: Soff es } Chief Marker: b _fY} 04, OF CRP. Signatures Internal Moderator: __/ G SrTpAVes Signature: Pia Si “a 4 DMCM: Moe WAH UE KE S Signature: Oe ny ba _ MCM: M TAM 80 Xt Signature: yb Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE 2 ENGINEERING SCIENCE N3 QUESTION 4 1.4 e In an elastic collision both momentum and kinelic energy are conserved. “ e In an inelastic collision only momentum ts conserved. ~~ 1.2 MyUy + MgUg = MV, + Moz 2x44+5x(-7) = 2x345xm,~ v= ~6,6 m/s” to the left” 1.3 1.3.1 = AY voutat OR a= re 20 = 0+ ax30” =e y 30-0 a = 0,667 m.s7? “ a= 0,667 ms7?2 1.3.2 Distance = 5bh + ixb OR Distance = Sat? + vt =0,5 x 30x 20+ 30x20”™Y~ =0,5x 0,667 x 307 + 20x30 = 900 m~ = 900m 14 1.4.4 Ti _3 > 1 Ty =v = 466,667 NY 142 v= aA(D+t)N v= 2x (0,06 + 0,005)x 5” v = 1,021 m/sv 1.4.3 P=(T-T.)xv P = (1 400 — 466,667)x 1,021 P = 0,953 kW (3 x 2) Copyright reserved Downloaded from hlayiso.com (2) (3) (2) (3) (8) [16]

SCIENCE W3. QUESTION 2 2.4 fi is the product of the perpenciculer force and the distance of the force from the fulerum. “~~ OR Turning point around the fulcrum “~~ (2) 2.2 2.2.1 A= 24,313 kNV D= 64,687 KN“ No mark for incorrect values for A and D (2) 2.2.9 a 24313-14313 _ 34687-19687 AB = ar ae cD = — = 5 kNim” = 5kN/m ¥ (4) 2.2.3 i OkN ¥ 14 kN V¥ 30 kN ¥ 5 kNim ¥ LLL ENOL i Y be A 2m Bl 3m Cl 3m D 4m Ey! | | 24,313 KNV 64,687 kNV 1 mark for supports A and D (1/2 mark each) (1) 1 mark uniformly distributed loads (1/2 mark each) (1) % mark for each point load (3 x 4) (1%) % mark for correct distance between point A to E (A) [12] Cepyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE Ae ENGINEERING SCIENCE NG QUESTION 3 3.4 3.2 3.3 3.4 They have the same magnitude” but are opposite in direction. or Equillbrant is balancing “and Resultant is replacing ~~ > Fup = s Faown Xsin45° + Ysin30° = 500 NY 0,707X + 0,5Y = 500 v ¥ Frignt = Y Fert Ycos30° = Xcos45° Y=0,816X v 0,707X + 0,5 x 0,816X = 500 1L115xX = 500 X= 448,43 NY And Y = 0,816(448,43) = 365,919 NY e The sum of all the forces in any direction must be zero. ” e The sum of all the moments of the forces about any fulcrum must be zero. 7” o = 13,75 0 BC “ and tan30 =—Eo «OF Cos30 = tas 13,75 sin30° = AB AB = 27,5kN ¥ BC = 23,816kNV BC = 23,816 kNV Strut ¥ Tiev Tiev Copyright reserved Downloaded from hlayiso.com (2) (6) (2) (4) [14]

MARGE DELINE 44 Prevents slip between two contacts surfaces V Removes material J Transmits movement by means of belts ¥ Brings moving machinery to a standstill Vehicle that cornering at a high speed {Any 3 * 1) (3) oo em ® 42 4.24 F, = 80 x 9,8cos30°” = 678,964 NY” (2) 4.2.2 f= 80x9,8sin30°~ =302N~ (2) 4.2.3 A, = 392N +. Constant velacity (1) 424 n= a or p=tanag p=tan 30° _ 392 = = s7e008” p=0,577 “ = 0,577” (2) 4.25 Fp = Fi, + Fy = 302 + 302” =784NY (2) [12) Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE -6- ENGINEERING SCIENCE N& OUESTICN § 5.4 It Is the quantity of energy required to raise the temperature of 1 kg of a substance by 1 °C. 4 (4) 5.2 5.2.4 E=Houxm = 30 Mu/kg = 0,75 kg = 22,5 MJ x 0,30 = 6,75 MJ (2) 5.2.2 E = mxshex(t2 - t,) 6,75 x 10° =6x 390(t, — 20) t, ~ 20°C = 2 B84,6°C fe = 2:904,6°C” (2) 5.2.3 AL=LoxaxAt =4x17 x 1075 x (2 904,6 - 20)” = 1,96 m~ Fiength = 4+ 1,96 =5,96mVv (3) 5.3 The temperature will increase with an increase in pressure. V OR The temperature will decrease with a decrease in pressure. V (1) 5.4 Hyer = Ag + qHypg 2600 = 743 + qx(2 030) “ q= 0,915 (2) 55 Ap =2x0,12 = 0,24 m2? AA = 0,240314 m? ~ 0,24 m? = 0,000314 m? AA = 2A,x @ xAt 0,000314 = 2 x 0,24 xax(70 — 15) “ @ = 1,189 x 1075/°Cw” (4) [15] Copyright reserved Downloaded from hlayiso.com

MARKING GU QUESTION € 64 « Total pressure V « Absolute pressure V e Gauge pressure « Pressure at a point 6.2 6.2.1 F f D2 2 — 70x 120 v 252 F=16128N” Fomg = 1612.8 |, 9,8 = 164,571 kg” ses F 622 p=£ OR pat A a = 70 wv we x(25 x 1073)2 4 P = 142 602,829 Pa OR 142,603 kPa 6.2.3 DexH=d*xhxan 120° x H = 25? x 80 x 20“ H = 69,4 mm Copyright reserved (Any 2 1) 1612,8 “x OAz2 P = 142,603 kPa Downloaded from hlayiso.com (2) (4) (3) (3) [12]

MARKING GUIDELINE -& ENGINEERING SCIENCE K3 QUESTION 7 74 It is the potential difference across the terminal of a flowing. ¥ 7.2 7.2.41 l= v 7 ~8Y 40. =2Aw 7.2.2 Vv Rey = ity 2A =50v 7.3 7.3.4 Q=It =8x2x60x 60” = 57600C” 7.3.2 m = Q x ece m = 57 600 x 0,001118 m = 64,49” 7.4 741 Np _ Is Ns Ip Np 82 Ns 3 Ratio 1:15 742 Np Is Ns Ip 300 = 0,2 y N, 3 N, = 4500” Copyright reserved cell when no current is 1) (2*2) (4) (2x2) (4) (2*2) 4) [13] Downloaded from hlayiso.com

QUESTION & 8.4 itis an Kis an 8.2 oe 8 @ 8.3 Copyright reserved “Ge ENGINEERING SCIENSE N3 electrically charged atom Vv OR atom that has gained or lost an electron. V Pure chemical corrosion V Surface corrosion ¥ Inter-granular corrosion ¥ Pure electrochemical corrosion (Any 3 « 1) Easily machinable ¥ Tough Vv Corrosion resistance Can undergo heat treatment Good conductor of electricity {Any 2 x 1) TOTAL: Downloaded from hlayiso.com (1) (3) (2) 6] 100

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higher education & i training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA T640(E)(N19)T NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 (15070413) 19 November 2018 (%-Paper) 09:00-12:00 a Please turn over Corrrantrssene! Downloaded from hlayiso.com

(15070413) -2- T640(EYN19)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions careful 3. Number the answers according | to the numbering system | used i in this question paper. « See oe : 4, Answers must be rounded off to THREE decimal places, 5. S ALL the calculations shi uld consist of at least the following THREE steps: ; The formula used: rthe manipulation thereof 6. 7. be used wherever possi 9. Keep subsections of ques ions together. 10. Rule off on completion of each question. 1. Use g = 9,8 m/s? 12. Write neatly and legibly. Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -3- T640(E)(N19)T QUESTION 1: MOTION, ENERGY AND POWER 1.1 State the law of conservation of energy in words. (2) 1.2 A motorcar with a mass of 0,9 ton is parked on an incline AB of 1: 20. The motorcar departs from rest at A and within 12 seconds reaches a speed of 144 km/h. If all frictional iosses are neglected, calculate: W =0.9 ton acceleration of the carrom A to B 1.2.2 1.2.3 1.2.4 1.2.5 (6x2) (10) 1.3 A Gautrain experiences a tractive resistance of 280 N per 1 000 kg when it moves. If the Gautrain exerts a force of 80 KN when it has a mass of 40 000 kg, calculate the acceleration of the Gautrain. (4) [16] Please turn over “opyrahtresened Downloaded from hlayiso.com

(15070413) 4. TS40{E\N19)T QUESTION 2: MOMENTS 2.1 State the law of moments. (2) 2.2 Given: a light horizontal beam ABCDEF of a uniform cross-section, loaded as shown. 50 KN 20 kN 6 kN/m 10 kN/m ULLEDEEDI TNT UML LTE SUL TELEITUNVESNNINEL ILA UIIITEREITET 2m + ims 3m 2m f 3m ry t i i : i : \¢ a 5 be ey sles » ' A © B co DOE F 224 Calculate the reaction of the supports B and E (4) “2.2.2 Draw a-Shear-force diagram, using a suitable scale. Show all the main values on the diagram (6) [12] QUESTION 3: FORCES 3.1 Explain the law. (4) 3.2 Determine the magni ude direction of the resultant force acting on the system of forces in equill im as shown in the figure below. 120 N 50N | 1 | | | | | | | \ 130 N (7) Copyright reserved Please turn over Downloaded from hlayiso.com

(18070413) -5- TS840(EKN19)T 3.3 Define the term direct force, and give TWO types of direct force. (4) [15] QUESTION 4: FRICTION 44 Give TWO disadvantages of friction as related to engineering. (2) 4.2 A body with a mass of 60 kg is placed onan inclined plane making an angle of 20° with the horizontal. The coefficient of friction is 0,4. A force F at an angle of 30° to the plane is required to pull the body up the plane. Calculate the following: .. : 4.2.1 The component F parallel to the plane (1) 4.2.2 The gravity component parallel. to the plane. SS (2) 4.2.3 6 The gravity component perpendicular to the plane (2) 424. The magnitude of F (4) 8 ar QUESTION 5: HEAT 5.1 Define ance. (2) 5.2 45 steel sha c : ‘litres of water. The initial temperature of the shafts is ater is 20 °C. if the final temperature of ixture is 93,97 °C, the specific heat capacity of steel is 0,5 ku/kg.°C of water is 4,187 kJ/kg.°C: Calculate the mass of EACH steel shaft. (3) 5.3 Steam is generated at a pressure of 2,85 MPa to a dryness fraction of 0,96. The steam is now superheated to a temperature of 300 °C. The specific heat capacity of this superheated steam is 2,9 kJ/kg.°C. Calculate, for 1 kg of steam, the amount of heat energy required for each of the following: 5.3.1 Enthalpy in the wet steam (1) 5.3.2 Dry steam (hg) (1) §.3.3 Superheated steam (hsu) (2) Please turn over Copyrantreseve! Downloaded from hlayiso.com

(15070413) -6- T640(E)(N19)T 5.4 A fuel with a heat energy value of 42 MJ/kg is used in an engine that develops 12 kW during a test. The test lasts for 35 minutes and a thermal efficiency of 45% is obtained. Calculate the following: 5.4.1 The equivalent power of the fuel 6.4.2 The heat energy dissipated by the fuel 5.4.3 The mass of the.fuel‘used: .. (3 x 2) (6) [15] QUESTION 6: HYDRAULICS 6.1 Define the unit pascal. | (1) 6.2 The following data refer to a single-acting hydraulic pres a Area of plunger = 0,09 m? Stroke of plunger = 13cm “Area of ram = 0,45 m2 © ~ Force applied to plunger = 300 N jnore any losses. {itres, displaced by the plunger in 10 y the ram, in mm, after one pumping stroke (2 * 3) (6) (2) 6.3 State Pascal's law in words 6.4 Indicate whether the following statements are TRUE or FALSE. Choose the answer and write only ‘true’ or ‘false’ next to the question number (6.4.1-6.4.2) in the ANSWER BOOK. 6.4.1 The SI unit of pressure is Pascal 6.4.2 1m? = 10 000 cm® (2* 1) (2) [11] Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -7- T640(EMN19)T QUESTION 7: ELECTRICITY 7A State Faraday's first law of electrolysis. (2) 7.2 Four cells, each with an EMF of 2 V and an internal resistance of 0.2 Q per cell, are connected in series. The battery is then connected to two parailel- connected resistors of 3 O and 6 O respectively, and a third resistor of 2 Q is connected in series with the two parallel resistors. Calculate the following: 7.2.4 The total resistance of the circuit (3) 7.2.2 The total current flow (2) 7.2.3 The voltage drop across the internal resistance (1) 7240 The current through the 3-Q resistor : : . a 5 (2) 7.3 The resistance of the heating ele a heater is 55 0. A current of 5A flows through the heater for 6 hours. (2 x 2) (4) [14] (1) (4) [6] TOTAL: 100 Conmigntreseve’ Downloaded from hlayiso.com

(15070413) ot TE40(E)(N19)T FORMULA SHEET All the formulae needed are not necessarily included. Any applicable formula may also be used. W=F.s my .uzptmy up =my Vv; tiny .V> Wepv D,=(D+) Ww P= 7 . Pravivet = hy + wg =? = Foy __ Vitset/Output 100% Palatn Paks Inset/Input P Fram — fpr A mb? F=ma Ae Apa APL 4 F, ; : Fram =Vp_xn R Aras ut = Apt Ley, p= tan® | By = eos Nr=Fot Fysina .. a=0 Fy = Fsin@ Fy = wsin® . Fo =weosd Fo =, -T; qT, . . —. = tension ratio T; P=F,.v ver.d.n W,, =F, 8 AL=Lo.a. At... Lp=Lo+ AL AE, =m.g.dh AA = Ay. B.A... Ap= Ag+ BA eye ny? AEx =S.m.av? 2a48=V u i 2 2 sS=u.d+—.at Q=I°.R.t 2 m=T1,z.t vrutat Vp _Np_Is STF=ax1F V5 Ng Ip M=F.1Ls Copyright reserved Please turn over Downloaded from hlayiso.com

(15070473) “2. SCH = SACHA Pass = Para + Per P= Sxgxh 1 i i =H Rear Ry R, Rseg = Ry +. Ry Vi-V, =—e(U,-U)) V=IxR Copyright reserved Downloaded from hlayiso.com T40(E)(N19)T

(15070413) INFORMATION SHEET T640(E}(N19)T PHYSICAL CONSTANTS CONSTANTS QUANTITY KONSTANTE HOEVEELHEID Atmospheric pressure 101,3 kPa Atmosferiese druk Density of copper 8 900 kg/m” Digtheid van koper Density of aluminium 2.770 kg/m” Digtheid van aluminium Density of gold 19 000-kg/m* Digtheid van goud Density of alcohol (ethyl) 790) kg/m’ Digtheid van alkohol (etiel) Density of mercury 13 600 kg/m Digtheid van kwik Density of platinum 21 500 ke/m* Digtheid van platina Density of water : 1 000 kg/m* | Digtheid van water Density of mineral oil 7 320 kgm eS “Digtheid yan minerale olie Density of ax 7,05 kg/m’ | Digtheid van lug Electrochemical equivalent of silver 1,118 mg/C Blektrochemiese ekwivalent van silwer Blectiochemical equivalent of CO} ber 0,329 mg/C Biektrochemiese ekwivalent van koper Gravitational acceleration 9,8 m/s” Heat value of coal <Warmiewaarde van steenkool '{;Warmtewaarde van antrasiet Heat value of petrol Warmtewaarde van petrol Heat value of hydrogen Warmtewaarde van waterstof Linear coefficient of expansion, of coppe Fx Lorre Lineére uitsettingskoéffisiént van koper Linear coefficient of expansion of aluminium “) 23 x 10°/°C Lineére uitsettingskoéffisiént van aluminium Linear coefficient of expansion of steel 12 x 10°C Line6re uitsettingskoéffisiént van staal Linear coefficient of expansion of lead 54 * 10°C Lineére uitsettingskoéffisiént van lood Specific heat capacity of steam 2 100 J/kg.°C Spesifieke warmtekapasiteit van stoom Specific heat capacity of water 4 187 S/kg.°C Spesifieke warmtekapasiteit van water Specific heat capacity of aluminium 900 Ikg.°C Spesifieke warmtekapasiteit van aluminium Specific heat capacity of oil 2 000 I/kg.°C Spesifieke warmtekapasiteit van olie Specific heat capacity of steel 500 J/kg.°C Spesificke warmtekapasiteit van staal Specific heat capacity of copper 390 I/kg.°C Spesificke warmtekapasiteit van koper Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUGLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 19 NOVEMBER 2018 This marking guideline consists of 8 pages. a i Chief Marker: AP NELE MA sionetuier Chief Marker: (hrsonene orp swan hip Internal Moderator Brocaghy AT Signature: fe eg) UMALUSI: LAKHAN NS Signature: Sood owew lV 4 Ly CE a onal ; wom: Vintaméo Xi seruue ill) Le 4 Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE t YO4QE}KNISYT ENGINEERING SCIENCE NZ QUESTION 1 44 Energy cannet be created or destroyed but can be changed from one form to another, (2) 4200 42400 Guat t _ 40-0 a= 73 a = 3,333 m.s-?¥ 1.2.2 F=ma = 900 x 3,3334 = 2999,7 NV 1.2.3 vy? =u* + 2as or Suit viou? =f LT ee (RONG Ly fae s=— s=Ox 12+ GSx (3,938) x (42ye 407-9? _- — s= deaaa ga 239 o7Tém ¥ s = 240,024 mv 1.2.4 _ Ex Pet P= 0,5(900)(40)? /(12)¥ P= Wai time = 2809,7_x 240,024” 12 P= 60000 W or 60 kWv P = 60 000 W or 60 kw 1.2.5 _ 4 1 ) _ > ace 6 = tan 50) = 2,86 Ep = mghsin® h= 240,024/ 20 = 12mv¥ E, = 900x 9,8x240,024xsin2,86°V Ey = 105629 J or 105,629 kj (5x2) (10) 1.3 Tractive resistance = 280 x 40 = 11 200 NY Fe=mxa R~ i, =ma 280 = 1000 xa 80 000 N — 11200 = 40 000x av'¥ ay = 0,28 mis* ¥ — 68800 Fa= " = Don a=mxa 80 G00 = 40 000 xa a@=1,72ms?v apt omis’ « . B= ap ~a, = 1,72 mis’ 4 (4) [16] Copyright reserved Downloaded from Please turn over hlayiso.com

MARKING GUIDELINE S: TOSHEYIN AGT ENCHNEERING SCIENCE NS QUESTION @ 21 A system of forces is in equilibrium if the sum of the left-hand moments about the fulcrum is equal ~ to the sum of the right-hand moments about the same fulcrum. (2) 2.2 2.2.1 Moment about B Yiacw =) cw 12x1 + Ex6 = 6x0,5 + 50x1 + 20x4+ 50x6,5¥ B= SS = 74,333 kNY Moment about E Siacw =)" cw Bx 6 + 30x1,5 = 20x1 + 20x2 + 50x5 + 6x5,5 + 12x7¥ Bz 382 = 63,667 KNY 6 (4) 22.2 ¥51,667 x 45,667 kNV 30 KNY -4,333 KN -12EN -24,333 kN Any six ~44,333 KNY (6) [12] Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE oh TO4QCE MAST ENGINEERING SCIENCE NS QUESTION 3 3.4 lf three forces acting on a point are in equilibrium, ~’then they can be represented in magnitude and direction” by the sides of a triangle” taken in order, ¥ 3.2 y Hy = 120cos45° ~ 50cos60° — 60V = —0,147N; West” DW = 130 ~ 50sin60° — 120sin45° v = 1,846N, North’ at -» (846 = tan \=0147 R= JOM? +1846 R =1,852N ; 85,459. ) = —85,45°¥ 3.3 This is a force which acts directly” and perpendicularly onto an object.“ Types: Tensile or pulling force” Compressive or pushing force” QUESTION 4 41 e Causes surfaces to wear away. e Drilling and cutting tools get blunt. * Friction causes unnecessary heat. (Any 2 x 1) 42 421 — F= Feos30°V or F = 0,855F 422 % =60 x9,8sin20°v F, = 201,108 NY 423 F, = 60 x 9,8cos20°V F, = 552,539 NY (4) (7) (4) [15] (2) a) (2) (2) Copyright reserved Please turn over Downloaded from hlayiso.com

ENGINEER! (4) [11] (2) (3) (1) (1) (2) 424 Ct o 7“ 2, fae * 2D Feown FB. + Fsin30° = 552,539 N Fy = 552,539 ~ 0,57” > Fright = » Frege Fcos30° = A, + 204,108 N Ff, = 0,866F — 201,108 NW F, a ose 0,866F — 201,10BN v 552,539 — 0,5F 221,06 ~ 0,2F = 0,866F — 201,108 1,066F = 422,124 F = 395,989 NY QUESTION 5 5.1 The specific heat capacity of a substance is the quantity of energy required to raise the temperature of kg” of a substance by 1° Celsius W 5.2 Hgainea = Mose 5x4187x(93,97 — 20) =m x 500x(180 °C — 93,97 °C)” 1548561,95 = mx 43015 m= 36,001 v mass of each steel shaft = 0,8kgv or 800 grams 5.3 5.3.1 Ayer = 995 +0,96 x 1807 = 2729,72 kJ /kgv §.3.2 H, = 2802 kj/kg¥ 5.3.3 Hey = Hy + Cyst = 2802 + 2,9(300 °C — 231°C)” = 3002,1 ki /kgv Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GLEDEL INE Ge. ENGINEERING SCIENCE NZ VESQELNTS)T . nae Pp 5.4 5.4.4 y= eh x 100 BP; Pin = 12 kw /0,45¥ Py = 26,667 kWY 5.4.2 E=Pxet = 26667 x 35x60¥ = 56 MJv 5.4.3 E=Hvxm m = 56 MJ/42 MJ¥ m = 1,333 kgv (3 * 2) (6) [15] QUESTION 6 6.1 The Pascal is a force of 1 N applied evenly and perpendicularly to an area of mv (1) 6.2 6.2.1 V=AXH = 0,09 x 0,13” =0,0117 m3 x 10 = 0,117 mov =117 iv 6.2.2 Aram * H = Aptunger XA 045 x H=0,09 x 0,13¥ H = 0,026 mv = 26mmv (2 « 3) (8) 6.3 In a closed system,” the pressure on the surface of a liquid is transmitted with the same intensity throughout the liquid in all directions. “ (2) 6.4 6.41 True 6.4.2 False : (2* 1) (2) [14] Gopyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE QUESTION 7 op ENCINEERING SCIENCE N3 TEGO NIE)? 7.4 The mass of a substance released or deposited” during electrolysis is propertional fo the current flowing through the solution. (2) 72 7.2.4 Rr = Ry + Rs + & 1 1 = f-+-y3¥v A ig + é +2402 x4 =4,80V (3) 7.2.2 I Vy r= Ry =8V/48 0" = 1,687 AY (2) 7.2.3 Veh xX tp V = 1,667 x 0,8 = 1,334Vv (1) 7.2.4 =I, x Rp = 1,667 x 2 = 3,334 Vw ¥; k= = = 3,334/3 = 1411 Av (2) 7.3 7.3.1 E=/F*Rt or E=PRt = 5? x 55x6x60x60Y =52x 55x6 = 29,7 MJ ¥ = 8,25 kW.h 7.3.2 27,45 27,45 = a = me cost = 29,7 MJ x 3eMy OR cost =8,25 x Tkw hk = R2,26” = R2,26 (2 * 2) (4) [14] Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE QUESTION & TE4QENKNAG)T “fe ENGINEERING SCIENCE NG 8.4 Oxidation is the loss of electrons by an atom. (1) 8.2 e Electroplating e Painting the metal e Oil and grease « Galvanizing (4) 83 NaCl (1) [6] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA — T640(E)(J26)T NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 26 July 2018 (X-Paper) 4 09:00-12:00 - of water and steam (BOE 173, mmable calculators m This question paper consist né jon sheets and 2 formula sheets. Please turn over Copmmontreseved Downloaded from hlayiso.com

(15070413) -2- T640(E)(J26)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question papel a sopest ete, 4. Answers must be rounded off to THREE decimal places. ° 5. All the calculations should consist of at least the following THREE steps: The formula:used or the manipulation thereof 5.2 f the given data in the form 5.3 The answer with the correct Si-unit . 6. Drawing instruments s i 7. All drawings/diagrams s 8. The constant values, as they app ihe attached information sheets, should be used wherever possibl 9. Keep subsections of questions together. 10. Rule off on completion of each question. 11. Use g = 9,8 m/s”. 12. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com

(15070413) -3- T640(E\J26)T QUESTION 1: MOTION, ENERGY AND POWER 1.1 What is meant by the term inertia of an object? (2) 1.2 A car is travelling from point A to point B on a circular path with a radius of 140 m, as shown in FIGURE 1. (Hint: C = 2 xr) A FIGURE 1 Calculate: 1.2.1, Distance f om Ato.B (2) 1.2.2 , from AtoB (3) 1.2.3 The trav f rs velocity from A to B in3 minutes (2) 1.3 FIGURE 2 shows lasses suspen 1 ya rope that passes over a frictionless pulley. Mas: 8 kg. The masses are allowed to move freely. A B FIGURE 2 Calculate: 1.3.1 The acceleration and direction of the masses (4) 1.3.2 The tensile force in the rope Nea Copwishtreseved ~~ Downloaded from hlayiso.com

(15070413) QUESTION 2: A. T640(E)(J26)T MOMENTS 2.1 What is meant by the term concentrated load? (1) 2.2 A uniform beam is 20 m long and rests horizontally on two supports. The one suppol rt is at the left end and the other support is 4 m from the right end. The beam carries concentrated loads of 200 N, 100 N and 80 N respectively 5 m, 10 m and 12 m from the left end. The beam carries a load of 15 N/m over the first 8 m from the right end. 2.2.1 Draw the diagram indicating all forces acting on the beam. (2) 2.2.2 Calculate the reactions at the supports. (4) 2.2.3 Draw the shearing force diagram. (4) ee [11] QUESTION 3:.FORCES 3.1. Name TWO conditions for a number of coplanai ‘point to be in equilibrium. _ (2) 3.2 A system of forces is shown in FIGURE 3: 2,5 kN ; FIGURE 3 Calculate: 3.2.1 The sum of the vertical components and state the direction 3.2.2 The sum of the horizontal components and state the direction 3.2.3 The magnitude of the equilibrant (4 x 3) (12) [14] Copyright reserved Downloaded from hlayiso.com

(15070413) 5 T640(E)(J26)T QUESTION 4: FRICTION 41 State THREE principles of kinetic friction. (3) 4.2 A casting with a mass of 500 kg rests on a surface with a slope of 20° to the horizontal. The coefficient of friction is 0,25. Calculate: 4.2.1 The component of the weight of the object perpendicular to the surface S (2) 4.2.2 The component of the weight of the object parallel to the surface (2) 4.2.3 The friction force between the casting and the incline (2) 4.2.4 The smallest force required to move the casting down the incline (2) QUESTION 5 HEAT 5.1 “Deseribe the effect of a change in pressure on the saturati r temperature. () 5.2 State TWO disadvantages | (2) 5.3 Water is h m 25°C to 75°C in 60 minutes by usi kg of a fuel with a heat value ; Calculate: 5.3.1 Change in (1) 5.3.2 Heat energy delivere he fue n 60 minutes (3) 5.3.3 Volume of the water ‘heated (4) 5.4 The sides of a square copper plate are 150 mm at 127 °C. The coefficient of neat expansion is 20 x 10°/°C. The temperature of the plate is decreased to Calculate: 5.4.4 Coefficient of area expansion (1) §.4.2 Final area (4) [16] Copwiantresevet Downloaded from hlayiso.com

(15070413) -6- TE40(EYKJ26)T QUESTION 6: HYDRAULICS 641 The pressure exerted at the bottom of a column of liquid is 30 kPa. The height of the column is 3,875 m. What type of liquid is used? (2) 6.2 Make a neat, labelled sketch to illustrate the application of Pascal's law. (3) 6.3 The following data refers to a single-acting hydraulic press: Force exerted onaram piston = 5200N Force applied to a plunger = 346N Diameter of plunger piston © = 0,08 m Stroke length of plungerpiston = 10 cm Calculate: : 6.3.1 Diameter of the ram piston (2) 6.3.2. - The distance. that the fam will tise Jn, 30 ) pumping strokes of the : plunger piston: (2) ee 6.3.3 The liquid pressure in the system 2s (3) 7.1 State Farad: d law of electrolysis. (2) 7.2 A battery has a « it.consisting of three cells, each with an EMF of 6 V and an internal resistanc nected in parallel. The battery is then connected to two resistors of 6 Q 4 Q which are connected in parallel and a third resist 0,4 Q connected in series. Calculate: 7.2.1 The total resistance of the circuit (4) 7.2.2 The current flowing in the circuit (3) 7.2.3 The potential difference of the parallel circuit (2) 7.3 An electric lamp is marked 230 V, 200 W. Calculate the cost to operate the lamp for 3 hours at a cost of 25 cents per unit. (3) {14] Copyright reserved ° Downloaded from hlayiso.com

(15070413) QUESTION 8: CHEMISTRY 8.1 What is meant by the term oxidation? 8.2 Name TWO methods used to combat corrosion. 8.3 Name the component elements of: 8.3.1 Brass 8.3.2 Table salt 8.3.3 Phosphor bronze Copyright reserved Downloaded from hlayiso.com T640(E)(J26)T (1) (2) (3 * 1) (3) [6] TOTAL: 100

(18070413) “te T640(E)(J26)T FORMULA SHEET All the formulae needed are not necessarily included. Any applicable formula may also be used. W= Fis my uy Eanty ly = Vp Ly V2 Wep.V D,=(D+ Ww P= ~ Fngtaver = hy + xdrjy Pe2adln..T=Fr _ Vitset/Output Inset/Input 100% 2 pafrim Fre 4.7? Aram Apt 4 Fy = Fcos0 Na =Fo + Fpsina a =0 Fy = Fsin@ Fy = wsin6 ee ZF y = Fycos0, + Fo =weos0 SF, = Risin +. + Fysin®, Freosa = Fy, + Fs... a=0 F.=1,-T, 2Fy tang = T —L = tension ratio * L Q=m.c.dAt..tp =lor At P=F,.v m.ww = QO = m.hy N ver.d.n..n=— 60 p-2 t Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -2- T640(E)(J26)T We, =F, oS AL=Lo.g. Mt... Ly= Lox AL dE, =n.g.Ah Ad= Ay. fh. At... Ap = An + SA i > yap oa AE, =5 may 24.8=V —-U i 2 s=u.+—al O=F Rut 2 m=I].z.t venta Vp Np _ Is a ZTFaIVE Ve Ns Ip M=F.is © CWM = SAC} iM Paps = Pary +-Pugr Copyrigntreseved = Downloaded from hlayiso.com

(18070413) INFORMATION SHEET PHYSICAL CONSTANTS T64AQ(E}(J26)T CONSTANTS KONSTANTE QUANTITY HOEVEELHEID Atmospheric pressure 101,3 kPa Atmosferiese druk Density of copper 8 900 kg/m Digtheid van koper Density of aluminium 2770 kg/m Digtheid van aluminium Density of gold 19.000 kg/m? Digtheid van goud Density of alcohol (ethyl) 790 kg/m Digtheid van alkohol (etiel) Density of mercury (600 kai | Digtheld van kwik Density. of platinum 21 “500 kein? Digiheid van platina Density of wales j 1 000 kg/m? Digtheid va water Density of mineral oil, 920 kg/m? Dietheid van tinea olie Density of air 1,05 kg/m* Electrochemical equivalent of silver Elektrochemiese ekwivalent van silwer Electrochemical equivalent of copper Elektrochemiese ekwivalent van koper Gravitational acceleration Swaartekragversnelling Heat value of coal Warmtewaarde van steenkool Heat value of anthracite 35 Mi/kg ‘Warmtewaarde van antrasiet Heat value of petrol 45 MJ/ke Warmtewaarde van petrol Heat value of hydrogen 140 MJ/kg Warmtewaarde van waterstof Linear coefficient of expansion of copper | 17 x 10°/°C Lineére uitsettingskoéffisiént van koper Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -2- T640(E\J26)T Linear coefficient of expansion of 23» 10°7°C Lineére uitsettingskoéfMfisiémt van aluminium aluminium Linear coefficient of expansion of steel 12 10°/°C Line€ére uitsettingskoéffisiént van staal Linear coefficient of expansion of lead 54% 10°C Line€re uitsettingskoéffisiént van lood Specific heat capacity of steam 2 100 J/kg.°C Spesifieke warmtekapasiteit van stoom Specific heat capacity of water 4.187 I/kg.°C Spesifieke warmtekapasiteit van water Specific heat capacity of aluminium Spesifieke warmtekapasiteit van Specific heat capacity of copper 900 kere aluminium Specific heat capacity of oi} 2 000 Vkg.°C Spesifieke warmtekapasiteit van olie Specific feat capacity of steel a 5007 kere - : : : Spesifieke warmtekapasiteit van staal 390 Jikg.°C Spesifcke warmtekapasiteit van koper Copyright reserved ~Downloaded from hlayiso.com

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ner edu Department: Higher Education and Training REPUBLIC OF SOUTH APRICA AMENDED MARKING GUIDELINE ENGINEERING SCIENCE N3 29 MARCH 2018 This marking guideline consists of 6 pages. Chief Marker: Ppnere Wh signature | i Chief Marker: i Of SORG AE =) nal Signature: Lie ~ Chief Marker: Boush PI Signature: /pi ae O/ _—_ Internal Moderator: SLAVE LG Signature: : DMCM: We HAH LELE q . Signature: MCM: Muranbo XH Signature: Copyright reserved D A al ae dl 2 dl f ann hh | ayi s0.co mi hii turn over

MARKING GUIDELINE TELOE)(M2O)T “2 ENGINEERING SCIENCE NS QUESTION 7 44 A vector quantity has magnitude and direction” while a scalar quantity has only magnitude. v (2) 4.2 4.2.4 S =ut + % at? =(0)(25) + %4(2)(25)°~ = 625 mv (2) 1.2.2 V=utat =0+ 2(50) =50 misY” S=vt = 50(2 « 60)¥ = 6000 mv (3) 1.2.3 Total distance = 625 + 6000 + 100 = 6725 mY (1) 1.2.4 V? = u? + 2as a=v-u? “Bs = (0? - 507V2(100)”” = -12,5mis*v Veutat t= (v- uj/a = (0 — 50)/-12,5 = 4sv Total time = 25+ 120+4 = 149sv (4) 1.3 Ex= Epv or Exs %mv* % mv*= mgh Ex= %x 5000 x (25) vv ¥(25)"= 9,8 x hv Ex 1,562MJ ¥ h = 31,89 mv Ep = EK S = h/Sin 10° Ep = 1,862 x 10° = 31.89/Sin 10° 5000 x 9.8 xh = 1,562 x 10% = 183.647 m ¥ h=3188m Vv 4) § = h/Sin 10° = = 31.89/Sin 10° [16] S = 183.647 mv Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE QUESTION 2 ERGINEER SCIENCE NS 2.1 € It is a force pushing one part of 2 body in one direction and another pert of the body in another direction. « It is the algebraic sum of all the forces on a particular side of a section. TO40(E (M22) 7 (1) (6) (4) [11] (2) (3) (3) (2) [10] (Any 1 * 7) 2.2 2.2.1 Moment about B ACM = CM Dx7+6x1+20*2 =16 *2+10 4 +25 x BV 7D = 272-46 D = 32,286kNY Moment about D YACM = YCM 10% 3+16x5+6x8+20x9=Bx7+25x1Vvv 7B = 338-25 B = 44,714 kNv 2.2.2 25y 18,7146 2.7144 ¥ Shape -207 ~7,286 ¥ -26 ¥ QUESTION 3 3.1 It is a single force that can replace two or more forces and still have the same effect. 3.2 3.2.1 Hx = -40 - 30cos45° + 60cos60°v ¥ = -31,213 Nv or (West/Left) 3.2.2 Vy = 50 - 30sin45 -G0sin60°v ~ =-23,175 NY or (Down/Souith) 3.2.3 R= jit? +¥,? = (31,213) + (—23,175)°v = 38,876 NW Copyrightreseved = Downloaded from hlayiso.coni™"""™

MARKING GUIDELINE -4- TE4O(E M2E}T ENGINEERING SCIENCE N3 QUESTION 4 4.4 It is the angle between the resultant force and the normal force. (4) 4.2 4.2.1 F = Pcos60° (1) 4.22 F = Wsin@é =80 « 9,8sin30°” = 392 NY (2) 4.2.3 F = Weos30° = 80 « 9,8cos30°Y =678,96 Nv (2) 4.24 Fy = Weos30° — Psin60°” Fu = Pcos60°- Wsin30°v” HOS FU/FN 0,4 = (0,5P — 392)/(678 96 -0,866P)v 271,584 ~ 0,346P = 0,5P ~ 392” 0,846P = 663,584 P = 784,378 NY (5) 43 e Prevents slip between two contact surfaces e Removes material e Transmits movement by means of belts « Brings movement to a standstill e Enables movement (Any 3 x 1) (3) [14] QUESTION 5 5.1 e Steam can be used repeatedly, e Coal and water are reasonably cheap and readily available. e Asteam installation has a tong life. (Any 2 * 1) (2) 52 5.2.1 = PowPin 45% = 15/Pinv Pi, = 15/0,45 = 33,33 KW 2) 5.2.2 E=Pxt = 33,33 x 10° 45 x 60V =89,991 MJv (2) 5.2.3 E=hvxm M = E/hv = 89,991 MJ/42 MJv = 2,143 kgv (2) Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE ~5- ENGINEERING SCIENCE NS TSEO(E]M 5.3 5.3.1 Qs mh, = 300,75 x 064% = 13,5 Mjv (3) §.3.2 Q=mcAt 13,5 x 10° = 10 x 900(t, — 25°C)” te = 1525°Cv (2) §.3.3 AL = lox a At = 2% 23 x 10% x(iZ00)¥ = O07 m Final length = 2 + 6,6¢ = 2,50 mv (2) 13] QUESTION 6 6.1 Suction head = static head ~ delivery head (1) 6.2 6.2.1 P = pgh 345 x 10° = 1000 = 9,8 x hv” h = 35,2 mv (3) 6.2.2 W=PxVv = 345 x 10°x 25/60V'Y = 143,75 kv (3) 6.2.3 P=EA = 143,75 x 105/60V = 2,396 kW (2) 6.3 If pressure is applied on an enclosed system, the pressure will spread evenly” through the enclosed liquid (or fluid) and is the same throughout.” (3) [42] Copyright reserved Downloaded from hlayiso.coni° turn over

MARIONG GUIDELINE QUESTION 7 ao ENGINEERING SCIENCE NS 7.1 The mass of a given substance liberated at an electrode of a cell is directly proportional” to the quantity of electricity passed through the cell.” (2) 7.2 7.2.4 VRp = 1/34 146 = 100” 7.2.2 Ir = Ve/Rr = 6/10” =0,6 AY 7.2.3 Vekxr =06x 2” =1,2Vv¥ 7.3 7.3.1 Np/Ns=Is/lp Ns =(600 = 6)/0,4” = 9000” 7.3.2 Np/Ns = Vp/Vs Vs = (240 x 9000)/600Y” =3600 VY 7.3.3 Ps = Vs.Is =3600 « 0,4” = 1,44 hwy QUESTION 8 8.1 An ion is a charged atom. 8.2 s Lithium e Sodium e Potassium e Rubium e Cesium e Fromide 8.3 8.3.1 Harder 8.3.2 Softer 8.3.3 79% Copyright reserved =2+iv 6 Rp=2Q¥ Rt= Rp + R int + Rseries 22+ 405)+4+2v7 (4) (2) (2) (2) (2) (2) [16] (1) (Any 2 x 1) (2) (1) (1) (1) [6] TOTAL: 100 Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA T640(E)(M29)T_ NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 | \ (1507041 3) REQUIREMENTS: ©. Properties ater and steam (BOE 173) a Calculators may be used. or, This question paper consists of ind information sheet Copyright reserved Downloaded from hlayiso.com"** turn over

(16070413) -2- T640(E\M29)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION |. Answer ALL the questions. 2. Read ALL the questions carefully, 3. Number the answers according to the numbering sys used in this question 4. 5. 6. 7. 8. The constant values, as they apyer 0 on the attached information sheet, must be used wherever possible. % é 9. Keep subsections of questions together. 10. Rule off on completion of each question. 11. Use g = 9,8 m/s” 12. Write neatly and legibly. Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) 3 T640(E)(M29)T QUESTION 1: MOTION, ENERGY AND POWER 1.4 1.2 1.3 QUESTION 2: MOMENTS 2.1 2.2 State the difference between a vector quantity and a scalar quantity. A lorry accelerates from rest to 2 m/s’ in 25 seconds and travels in a westerly direction from point S to point T. From point T it then travels at a constant velocity for two minutes towards point U and then starts to brake and stops at point V, which is 100 m from point U. Caiculate: 1.2.1 Displacement from S to T. 1.2.2 Displacement from T to U 1.2.3 Total distance travelled 1.2.4 Time taken for the journey A lorry of 5 tons travels at 90 krn/h and switches off its ts engine when reaching an incline of 10°. . Calculate how far the lorry will travel up the incline if friction is not considered. What is meant by shear for 20 kN 25 kN 3 kN/m . t i 2m 4m ‘ : 1m _ vl ale A B Cc D E The figure above shows a light horizontal beam A, B, C, D and E of uniform cross section, loaded as shown. Calculate: 2.2.4 The reactions of the supports B and D. 2.2.2 Draw a shear force diagram using a suitable scale. The diagram must be fully labelled. Please turn over Copviantreseved = Downloaded from hlayiso.com (2) (3) (1) (4) (4) [16] (6) (4) [11]

(15070413) -4- T640(EKM29)T QUESTION 3: FORCES 3.1 Explain what is meant by the resultant of a number of forces. (2) 3.2 50N 4 150° 40N Calculat 2: 3210 | (3) 3.2.2 (3) 3.2.3 (2) [10] QUESTION 4: FRICTION 41 What is meant by angie of frictio (1) 4.2 A body of mass 80 kg is placed on an. incline’ of 30° with the horizontal .The coefficient of friction is 0,4. A force P makirig an angle of 60° with the plane is required to pull the body up the plane. Calculate: 4.2.1 The component of P parallel to the plane (1) 4.2.2 The gravity component parallel to the plane (2) 4.2.3 The gravity component perpendicular to the plane (2) 4.2.4 The magnitude of P (5) 4.3 State THREE advantages of friction as applicable in engineering. (3) [14] Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -5- T640(E(M29)T QUESTION 5: HEAT 5.4 Give TWO advantages of steam. (2) 5.2 A fuel with a heat value of 42 MJ/kg was used in an engine that develops 15 kW. The test lasted for 45 minutes and a thermal efficiency of 45% was obtained. Calculate: 5.2.1 Equivalent power of the fuel (2) 5.2.2 Heat energy dissipated by the fuel © ; (2) §.2.3 Mass of the fuel : (2) 5.3 The following data refers to an aluminium rod, 200 cm in length, being heated by burning coal: Initial temperature of the rod Amount of coal burned Percentage of heat t ansferred to the rod Mass of the rod : Calculate: 5.3.1 _The quantity of heat transferred to the rod : (3) §.3.2 Final temperature ofthe rod (2) 5.3.3. The final length of the rod (2) / os N, | [15] QUESTION 6: HYDRAULICS ' VON 6.1 Write an equation for a suct head in terms of delivery head and static head. \ : (1) 6.2 The following data refers to a single stroke pump, driven by a diesel engine, which has to deliver water to a reserveir. above'the pump: Working pressure of pump = 345kPa Volume of water delivered = 25 m%/hour Output power of the diesel engine. = 9,45 kW Calculate: 6.2.1 Maximum height of the reservoir above the pump (3) 6.2.2 Work done by the pump per minute (3) 6.2.3 Output power of the pump (2) 6.3 State Pascal's principle in words. (3) [12] Copyright reserved Downloaded from hlayiso.com’*** turn over

(15070413) -6- Te4Q(EKM29)T QUESTION 7: ELECTRICITY 7.1 State Faraday's first law of electrolyses in words. (2) 7.2 A circuit consists of four cells each with an EMF of 1,5V, connected in series. Each cell has an internal resistance of 0, 5 Q. Two resistors of 3 Q and 6 QO are connected in parallel to the circuit. The battery is now connected in series with a radio of 4 Q and a loudspeaker of 2 Q. Calculate: 7.2.1 7.2.2 7.2.3 7.3 A single-phase transformer, has a supply voltage of 240 V and a primary current of.6 A at full load. The ‘secondary curren 600 turns'on the pritr : Total resistance of the circuit Current flow in the circuit Voltage drop across the loudspeaker ry coil. | Calculate: : 7.3.4 : Number of turt én the secondary coil 7.3.2 Secondary voltage 7.3.3 Secondary pow: oo. QUESTION 8: CHEMISTRY \ : 8.1 Give the definition of an ion \ 8.2 Give TWO examples of alkali earth metals. 0, 4 A and there are 8.3 Complete the following statements by choosing the correct option in brackets. Write only the answer next to the question number (8.3.1-8.3.3) in the ANSWER Zine is (8.3.1 smoother/harder/softer) than tin but (8.3.2 smoother/harder BOOK. /softer) than copper. Nitrogen is a colourless, odourless and tasteless gas and forms (8.3.3 95% / 60% / 79%) of the total volume of the air. Copyright reserved Downloaded from hlayiso.com TOTAL: (2) (2) [16] (1) (2) (6) 100

(16070413) A T640(EM29)T ENGINEERING SCIENCE N3 FORMULA SHEET All the formulae needed are not necessarily included. Any applicable formula may also be used. W= Fs nay Uy Eig Uy = My Vy E12y. Vz W=p.V ° D, =(D +1) Ww ov Prarrvet = hy + xh fe f a | Peladlin.T= Fr n= Uilset/Output 100% : Jnset/Input pe Frau - Fer As xD F=ma : SO Aram Apr 4 F, : Vea =Vepx HT 4 & Nr Aram - Aran = Ary - ua=tan® Fy = Feos@... , Ng = Fo + Frsina ..a=0 Fy = Fsin0 Fy = wsin@ oSFy = Fcos0, + ... + F,cos6, Fo = weos@ SF, = Fysin6) +... + F,sin6, Freosa = F,, + Fy ..a=0 F.=1,-T, q . . — = lension ratio 2 P=FE,.v vea.d.n.n=—- 60 W,=F,.s AL =Lo.a. At... lp= Lo AL AE, =m.g.Ah AA= Ay. B. At... Ap Ag+ AA 4Ex =4.m.Av? 2asev =u" rae 2 seutttas O=l Rt 2 m=I.z.t veutast Vp Np _ Is ZTF=Z1F Vs Ns Ip M=F.1s Copyright reserved Please turn over Downloaded from hlayiso.com

(18070413) ~2- 2S CWM = SACHWM Pans = Pary + Puer Px dx gxh ] / 1 = tbo Rpg R; R, a Rspep= Ry +. Ry V,—Vy =-e(U;-Us) V=IxKR Copyright reserved Downloaded from hlayiso.com T640(E\M29)T

(15070413) INFORMATION SHEET PHYSICAL CONSTANTS T640(EYM29)T Gravitational acceleration 08 mie. CONSTANTS QUANTITY KONSTANTE HOEVEELHEID Atmospheric pressure 101,3 kPa Atmosferiesedruk Density of copper 8 900 kg/m? Digtheid van koper Density of aluminium 2770 kg/m" Digtheid van aluminium Density of gold 19 000 kg/m" Digtheid van goud Density of alcohol (ethyl) 790 kg/m* Digtheid van alkohol (etiel) Density of mercury 13 600 kg/m* ‘Digtheid van kwik Density of platinum 21 500 kg/m Digtheid van platina Density of water 1000 kg/m? Digtheid van water Density of mineral pil 920 kg/m? Density of air 7 1,05 kgm? —_ Electrochemical equivalent of silver 1,118 mg/C Electrochemical equivalent of copper 7 0,329 mg/C Swaartekragver: Heat value of coal 30 Mi/kg.. Warmtewaarde van steenkool Heat value of anthracite 35 Milks Warmiewaarde van antrasiet Heat value of petrol % Wanntewaatde van petrol Heat value of hydrogen Wanntewaarde van waterstof Linear coefficient of expansion of copper | Lineéreuitsettingskoéffisiént van koper Linear coefficient of expansion of Lineéreuitsettingskoéffisiént van aluminium aluminium Linear coefficient of expansion of steel 12x 10°C Lineéreuitsettingskoéffisiént van staal Linear coefficient of expansion of lead 54x 107°C Lineéreuitsettingskoéffisiént van lood Specific heat capacity of steam 2100 Ikg.°C Spesifickewarmtekapasiteit van stoom Specific heat capacity of water 4187 Ike Spesifiekewarmtckapasiteit van water Specific heat capacity of aluminium 900 J/kg.cC Spesifiekewarmtekapasiteit van aluminium Specific heat capacity of oil 2 000 J/kg.°C Spesifiekewarmtekapasiteit van olie Specific heat capacity of steel 500 J/kg.°C Spesifiekewarmtekapasiteit van staal Specific heat capacity of copper 390 Jkg.°C Spesifiekewarmtekapasiteit van koper Copyright reserved Downloaded from hlayiso.com

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Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA AMENDED MARKING GUIDELINE ENGINEERING SCIENCE N3 29 MARCH 2018 This marking guideline consists of 6 pages. rh? Chiet Marker; APNE LE WH signature: | Chief Marker: MY. oT SoagAe —DkD Signature: ~ Chief Marker: Brough py Signature: _ a, Internal Moderator: Sra EG ]}: G Signature: £ te DMCM: W\ensnrecé o 5 Signature: Klute MCM: My tame XH Signature: As Co Copyrightreseved = Downloaded from hlayiso.coni*""" 7"

3 as MARKING GUIDELINE Be TEA QMEKM2 ENGINEERING SCIENCE NS « QUESTION 1 1.1 A vector quantity has magnitude and direction” while a scalar quantity has only magnitude. ~ (2) 1.2 1.2.1 S=ut+ %at? =(0)(25) + 14(2)(25)? = 625 mv (2) 4.2.2 Veutat =0+ 2/50) =50 mis¥ S=vt = 50(2 x 60)v = 6000 mv (3) 1.2.3 Total distance = 625 + 6000 + 100 = 6725 mv (1) 124 V =u2+2as asv—u? 28 = (0? — 507)/2(100)¥ = -12,5mis’v Veutat t=(v-ujla = (0 ~ 50)/-12,5 = 4sv Total time = 25 + 120+4 = 149sv (4) 13 Ex= EpY or Ex=%mv* % mv"= mgh Ex= 4 x 5000 x (25)? vv ¥4(25)"= 9,8 x hv Ex=1,562MI ¥ h= 31,89 mv Ep = EK S = h/Sin 10° Ep = 1,562 x 10° = 31.89/Sin 10° 5000 x 8,8 xh = 1,562 x 10%) = 183.647 mv h=318o9m Vv (4) S = h/Sin 10° = = 31.89/Sin 10° [16] S = 183.647 mv Copyright reserved Please turn over Downloaded from hlayiso.com

TOZOLEMM2E)T we MARKING GUID ENGINEERING SCIENCE NG a QUESTION @ 2.1 « [tis a force pushing one part of a body in one direction and another part of the body in another direction. e It is the algebraic sum of all the forces on a particular side of a section. (Any 1 * 1) (1) 2.2 2.2.1 Moment about B SACM = CM Dx7+#6x%1+20*%2 =16 *2+10 * 4 +25 x B¥v 7D = 272 —- 46 D = 32,286kN~ Moment about D DACM = YCM 10*%34+16*5+6x8+20x9=Bx7+25x 1vv 7B = 338 ~ 25 B = 44,714 knw (6) 2.2.2 25. -18,714+ 7148 ¥ Shape -20¥ -7,286 V 26 V (4) [11] QUESTION 3 3.1 It is a single force that can repiace two or more forces and still have the same effect. (2) 3.2 3.2.1 Hx = -40 ~ 30c0s45° + 60cos60°v “ = 31,213 NY or (Wesi/Left) (3) 3.2.2 Vy = 50 — 30sin45 -60sin60°vY =-23,175 NY or (Down/Souilth) (3) 2.3 32 R= |i,’ + v4? = /(—31,213)? + (—23,175)2” = 38,876 NV (2) [10] Copyrightreserved Downloaded from hlayiso.coni®:""

MARKING GUIDELINE “An T6ANEMM29)T ENGINEERING SCIENCE N& QUESTION 4 4.4 It is the angle between the resultant force and the normal force. (1) 42 4.2.4 F = Pcos60° (1) 4.2.2 F = Wsine =80 x 9,8sin30°“ = 302 NV (2) 4.2.3 F = Weos30° = 80 x 9, 8cos30°v =678,96 NV (2) 4.2.4 Fy =Weos30° — Psin60°” Fu = Pcos60°- Wsin30°” Ho = Fu/Fy 0,4 = (0,5P — 392)/(678,96 -0,866P)v 271,584 — 0,346P = 0,5P ~ 392“ 0,846P = 663,584 P = 784,378 NY (5) 43 e Prevents slip between two contact surfaces « Removes material * Transmits movement by means of belts 2 Brings movement to a standstill e Enables movement (Any 3 = 1) (3) 14] QUESTION 5 5.1 ® Steam can be used repeatedly. » Coal and water are reasonably cheap and readily available. * Asteam installation has a long life. (Any 2 * 1) (2) 5.2 5.2.1 = Pow/Pin 45% = 15/P i.” Pin = 15/0,45 = 33,33 kWv (2) 522 E=Pxt = 33,33 x 10° x 45 x 60V =89,901 MJv (2) 5.2.3 E=hvxm M = E/hv = 89,991 MJ/42 MJ” = 2,143 kgv (2) Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING Gl TOA QEEYT ENGINEERING SCIENCE N& 6.3 6.3.4 Q= mi, = 300,75 x 06% = 13,5 Mjv (3) §.3.2 Q = mcht 13,5 x 10° = 10 x 900(t, — 25°C)” ty = 1525°CV (2) 5.3.3 AL=Lox ade = 2x23 x 10°x(! 2 O6Y m Final length = 2 + 5,69 = 2,5 mv (2) [15] QUESTION 6 6.1 Suction head = static head — delivery head (1) 6.2 6.2.1 P =pgh 345 x 10° = 1000 x 9,8 x hv h= 35,2 mY (3) 6.2.2 WePxV = 345 x 10°« 25/60V~ = 143,75 klv (3) 6.2.3 P=Er = 143,75 « 105/60” = 2,306 kWY¥ (2) 6.3 If pressure is applied on an enclosed system, the pressure will spread evenly” through the enclosed liquid (or fluid)“ and is the sarne throughout.” (3) [12] Copyright reseved = D)gwnloaded from hlayiso.coni*""™

MARKING GUIDELINE ENGINEERING SCIENCE N3 TEAO(EMZ9)T QUESTION 7 74 The mass of a given substance liberated at an electrode of a cell is directly proportional” to the quantity of electricity passed through the cell.” (2) 72 7.2.4 Rp = 1/3 + 146 =24+1V7 6 Rp=2Qv” Rt = Rp + R int + Rseries =2+4(05)+4+2v = 100” (4) 7.2.2 by = Vy/Rr = 6/10V =0,6 AY (2) 7.2.3 Vez lpr =0,6 x 2” =1,2 Vw (2) 7.3 7.3.4 N,/Ns=Is/Ip Ns =(600 « 6)/0,4~ = 9000” (2) 7.3.2 Np/Ns = Vp/Vs Vs = (240 x 9000)/600Y =3600 VY (2) 7.3.3 Ps = Vs.Is =3600 x 0,4” = 1,44 kwWY (2) [16] QUESTION 8 8.1 An ion is a charged atom. (1) 8.2 e Lithium ¢« Sodium »® Potassium « Rubium * Cesium (Any 2 * 1) e Fromide (2) 8.3 8.3.4 Harder (1) 83.2 Softer (1) 8.3.3 79% (1) [6] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA T640(E)(N20)T NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 (15070413) 20 November 2017 (X-Paper) 's09:00-12:00 REQUIREMENTS: « Properties of Water and Steam (BOE 173) Calculators may be used, This question paper consists of 7 pages, a formula sheet of 2 pages and : 1 information Sheet, Copyright reserved Downloaded from hlayiso.com"** turn over

(15070413) -2- T640(E)N20)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully: 3. Number the answers according to the numbering system used in this question paper. a gh es 4. Answers must be rounded off to THREE decimal places, 5. All calculations should consist of at least the following THREE steps: _ * The formula used or thé manipulation thereof . in the formula . 6. 7. 8. Rule off on completion of each ue: 9. Use g = 9,8 m/s’, 10. Write neatly and legibly. Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -3- T640(E}(N20)T QUESTION 1: MOTION, ENERGY AND POWER 1.1 State the law of conservation of momentum. 1.2 A body of mass 2 kg moving at 8 m/s collides from the rear with a 5 kg body moving at 5 m/s. After collision the 2 kg body continues to move in the same direction at 3 m/s. Calculate the new velocity of the 5 kg body. 1.3 A flat belt has a thickness of 7 mm. The belt fits on a pulley with a diameter of 0.494m. The rotational frequency of the pulley is 1 200 r/min. The effective force (Fe) transmitted by the belt is 250 N. :. Given that the belt width is 10 cm and the maximum force per mm width is 5 N, calculate the following: 1.3.1 The velocity of the belt in m/s 1.3.2 The power transmitted. =. 1.3.3. The maximum force (T:) in the tight side 1.3.4 The force (T2) in the slack side QUESTION 2: MOMENTS 2.1 Define the term moment of a force. 2.2 ry ry FIGURE 1 FIGURE 1 above shows a light horizontal beam ABCDE of uniforms cross- section, loaded as shown. 2.214 Calculate the magnitude of the reactions of supports B and E. 2.2.2 Draw the shear-force diagram. Copyantreseved ~— Downloaded from hlayiso.com (3) (2) (2) (1) [14] (1) (4) (5) {10} Please turn over

(15070413) -4- T640(E)(N20)T QUESTION 3: FORCES 3.1 What is meant by the term triangle of forces? (2) 3.2 1OOKN 40kN B E - LA 45° / 459 45° 4m A 4m ; FIGURE 2 ose - Calculate the following: 3.2.1. The reactions at the support P and Q © (3X2) 6) 3.2.2 he magnitude of BE and EA — (3x2) 6) : - a [14] QUESTION 4: FRICTION 4.1 What is meant by the ter | angle of rest? (2) 4.2 A steel block with a mass of 5 kg rests on a horizontal plane and a force of 12 N is required to set the block in. motion and a horizontal force of 9 N is needed to keep it in motion. . ~ Calculate the following: 4.2.1 The coefficient of static friction 4.2.2 The coefficient of kinetic friction (2x 2) (4) 4.3 An object is pulled upwards against a slope of 20° with the horizontal. A force of 300 N, parallel to the incline, is required to pull the object with a mass of 50 kg up the incline. Calculate the coefficient of friction between the surfaces in contact. (5) [14] Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) 5 T640(E)(N20)T QUESTION 5: HEAT 5.1 Define the term saturation temperature. (1) 5.2 A steel block with a mass of 1,5 kg is placed in an electric oven with a capacity of 10,8 MJ per hour. The temperature of the block rises from 25 °C to 800 °C. The specific heat capacity of the block is 0,503 kJ/kg. °C. Assume that only 68% of the heat reaches the block. Calculate the time taken to raise the temperature of the block. (4) §.3 An engine has a brake power of 16,2 KW. The engine is subjected to a brake test for 40 minutes. The mass of the fuel consumed is 2,25 kg and the heat energy of the fuel is 32 MJ/kg. Calculate the following: 5.3.1 The heat energy released by the fuel (2) 5.3.2 The equivalent power of: the engine. oh ee : (3) §.3.3 The efficiency of the engine ae S {2) 5.4 Complete the table below by writing down the missing values (a—c), without referring 4o the steam tables, next to the question number (S. 4) in the ANSWER F BOOK. a _H; (kJ/kg) Hey (kdikg) : co (kJ/kg) a 318 2319 | () b 492 ra) 2702 c (1) 2117 2744 (3) [15] QUESTION 6: HYDRAULICS 6.1 With reference to hydraulics, wha’ ‘4s meant by the term pressure? (1) 6.2 The ram of a hydraulic jack is 50 mm in diameter. The diameter and stroke of the plunger are 20 mm and 50 mm respectively. The mechanical advantage is 25. Determine the following: 6.2.1 The force that must be applied to the lever to lift a load of 2 200 kg if the efficiency is 75% (5) 6.2.2 The number of strokes of the lever to lift the load 12,6 cm (2) copyightreseved ~~ Downloaded from hlayiso.com’*"~

(15070473) -6- T640(E)(N20)T 6.3 The following data refers to a single-acting hydraulic press: Area of plunger = 0,09 m? Stroke of plunger = 13 cm Force required to plunger = 0,3 KN Area of the ram = 0,55 m* Ignore any losses. Calculate the following: 6.3.1 The volume of the liquid displaced by the plunger in 10 strokes 6.3.2 The distance moved by the ram in mm, after one pumping stroke of the plunger : (2 * 3) (8) [14] QUESTION 7: ELECTRICITY 71 State Joule's law in words. keel : oe (3) 72 40 2 / 3Q FIGURE 3 Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) “7 T64Q(E)(N20)T Three cells, each with an EMF of 2 V are connected in series with an external circuit as shown in FIGURE 3 above. Calculate the following: 7.2.1 The total resistance of the circuit (4) 7.2.2 Reading on the ammeter (2) 7.2.3 Reading on the voltmeters V; and V2 (4) 7.2.4 Current through the 4 Q resistor itel QUESTION 8: CHEMISTRY 8.1 Give TWO properties each for metals and non-metals. (2 « 2) (4) 8.2 One of the precautions against corrosion is by protecting t the metal with a layer of plastic or fibre glass. : Give TWO conditions where plastic will not be suitable. oe (2) TOTAL: 100 Copmightreseved =~ Downloaded from hlayiso.com

(15070413) -1- T640(E)(N20)T ENGINEERING SCIENCE N3 FORMULA SHEET All the formulae needed are not necessarily included. Any applicable formula may also be used. We Fs Nt, Uj EM. = My Vp EM .V2 Wepv D,=(D+0) Ww P= 7 Pnarwer = Hye + XAyy . Ps2latin..T=F r - Uitse/Our put 100% ; Inset/Input P= Fran _ Fp ds nD F=ma : Aram — Apt, 4 path Fram = Vex 7 =. ® Aram. Upsit = Apis EPL f= tan® Fy = Feos0 =F. + Frsi a Ng =Fot Frsina...a=0 Fy = Fsind Fs = wsind SF = Fycos0, +... + F,cos0, Fo = weos? | ER, = Fisinby + ERwin R= SF? + EEF Freosa = Fi, + Fy .. d= 0. Fo=T,-T, tang = 7 = tension ratio ; 2Fy : ; -, Q=m.c.At..tp =tot At P=F,.v = ‘e° ven.d.nw n=’ 60 Wy = Fy. s AL=Lo.a. At... Ly=Lo® AL AE, =in.g.Ah AA= Ay. B. At... Ap= Ag AA 4Bg =4.m.dv? 2assv -u? 2 ! at QO=!I°.R.t sauit at m=l.zt veurtat Vp Np _ ds XTF=SVF Ve ON L sous? M=F.1Ls Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) -2- & CWA = S4ACIWAL Piss = Pare t+ Prey P= dx gxh I I I = 4.t Roar By Rx, 7 Rser = Rp + Ry Vj Vy =~ e(U) U3) V=IxR Copyrigntreseved =~ Downloaded from hlayiso.com TG40(EXN20)T

(16070413) INFORMATION SHEET PHYSICAL CONSTANTS QUANTITY CONSTANTS Atmospheric pressure 101,3 kPa Density of copper 8 900 kg/m” Density of aluminium 2770 kgim Density of gold 49 000 kgim* Density of alcohol (ethyl 790 kgim? Density of mercury 13 600 kgim= Density of platinum 24 500 kgim* Density of water 1000 kg/m Density of mineral oil 920 kgim? Density of air 1,05 kgin* Electrochemical equivalent of silver... = 1.118mgic Electrochemical equivalent of copper 0,329 mg/C Gravitational acceleration 9,8 mis” Heat value of coal: 30 MJ/kg Heat value ofan 5 35 MJ/kg Heat value of petrol | 45 Muikg™ Heat value of hydrogen : : 140 MJ/kg Linear coefficient of expansion of copper : 17x 10° PC Linear coefficient of expansion of aluminium Linear coefficient of expansion of steel” Linear coefficient of expansion of lead £84 x 10°C Specific heat capacity of steam 2 100 J/kg.°C Specific heat capacity of water 4 187 Jikg.°C Specific heat capacity of aluminium 900 Jikg.°C Specific heat capacity of oil 2 000 J/kg.°C Specific heat capacity of steel 500 J/kg.°C Specific heat capacity of copper 390 J/kg.°C Copyright reserved Downloaded from hlayiso.com T640(E})(N20)T

* education “al _ Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA / | | AMENDED MARKING GUIDELINE / NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 20 NOVEMBER 2017 This marking guideline consists of 6 pages. Chief Marker: 0-77, WA O7SORGCHE Signature: | 7 Chief Marker: Kish Raves ra Signature: ie — | ¢ Chiet Marker; VL Oaig Wied Signature: (Ey © Internal Moderator: TT g Srtmam € _ Signature: Gane UMALUSI Moderator: Nc LAKHAA Signature: Co, DMcM: Ne ud NUGUe 4 Signature: OA Lee, MCM: M NTAMBa Xt Signature: a Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING GUIDELINE . TAEOLE KIIOVT QUESTION 1 14 If no external force acts on a system of objects that collides, the tolal momentum will remain constant. (2) 1.2 Poetore = Patter 2kg x 8m.s71 +5kg X Sms" =2kg X 3m +54 41 kg.m/s — 6 kg.nv/s = § vo” V2 = +7 mis¥ (4) 4.3 1.3.1 Effective diameter = (D + t) = 0,494 + 0,007 = 0,501m” V=n(D+t)n V = Tr * 0,501 * 1200/60¥ = 31,479 mis¥ (3) 4.3.2 P=Fexv = 250 N * 31,479 m/s ¥ = 7869,75 WY (2) 1.3.3 T; = belt width x maximum per mm belt =5Nx 100V = 500 NY (2) 1.3.4 F.=11-Tz Te =T1-Fe = 600 — 250 = 250 NY (1) [14] QUESTION 2 21 Product of perpendicular force and the distance of the force from the fulcrum“ OR The turning effect of a force about a point” (1) 2.2 2.2.1 Take the moment about E XCW =yACW Bx7 =10*2 +16*1+10*5+10*6+15% 85” B = 39,071 kNY Take moment about B DACW =SCW Ex7 + 15x 1,5=10%1+10*%2+10*5+ 16x 6” 7E = 153,5 E = 21,929 kKNY (4) Copyright reserved Please turn over Downloaded from hlayiso.com

MARKING CLUDELINE 3 TE4UERNZOYT ERNGIREERING SCIENCE NS -21,929KN scl one otark for the correct shape (5) [70] QUESTION 3 3.1 If three forces acting on a point are in equilibrium~, they can be represented in magnitude and direction by the sides of a triangle taken in order. ~ (2) - 3.2 3.2.1 Moments about Q sCW =TACW Px8=100x6+40x2v P = 680/8 ¥ P= 85 kN“ Moments about P SCw =yACW Qx8 = (100 x 2)+(40* 6) ¥ Q= 440/8 ¥ Q= 55 kNY (6) 3.2.2 85/BE = sin 45°“ BE = 120,21kNY Strut Concession of one mark given because the nature was not asked Tan 45 = 85/AE v AE = 85 kN ¥ Tie Concession of one mark given because the nature was not asked (6) Copyright reserved Please turn over Downloaded from hlayiso.com

YEO KNZO)T ENGINEERING SCIENCE NG i14} QUESTION € 44 The biggest angle that may be reached between an inclined plane and a horizontal plane - before a body starts moving freely down the incline. (2) 4.2 4.24 Us = F/N = 12/49Y = 0,245” 42.2 y= FUN = 9/49 ¥ = 0,184" (2*2) (4) 43 A, = F —Wsin@ = 300 — (50 x 9,8sin20°)~” F, = 132,410 NW NR = (50 x 9,8)cos20° = 460,449 NY = te u= NR = 132,410N 460,449 . = 0,288 ¥ (5) [41] QUESTION 5 5.1 The temperature at which a liquid boils at a fixed pressure “ OR The boiling point of a liquid (1) 5.2 Hv= meat =1,5* 0,503 « 775“ = §84,738 = 0,585 Mu HVsteel —_0,585 H¥ oven — 10,8 x 68% = 0,0786 Mu T=0,0796 = 60 = 4,778 Minutes” (4) 5.3 5.3.1 E=hvxm = 32 x 2,26V = 72 Mv (2) 53.2 P=Er = 72 x 1082 400vv = 30 kWY (3) §.3.3 Efficiency = Pou/Pin ¥ 100 = (16,2/30) x 100” Copyright reserved Please turn over Downloaded from hlayiso.com

5.4 a) 2 637 ) 2244 ©) 627 QUESTION 6 6.1 Force per unit area“ 6.2 6.2.1 FID? = tid’ f= (21 560 20°50?” = 3 449,6 NV E=f/MA = 3 449,6/25 = 137,984 NY“ Actual force required = (137,894 x 100)/75” = 183,979 NV 6.2.2 No of strokes = (D*H)/d"h = (50° x 126)/207 x 50)” = 15,75 = 16 strokes” 6.3 6.3.1 VEAxh = 0,08 x 0,13” = 0,0117 m? = 10” = 0,117 mv 6.3.2 A*H=A * stroke length 0,55 * H= 0,09 * 0,13” H = 0,021 mv = 21,273 mmv Copyright reserved oye & VEAUENNEOT (2) (3) [13] Q) (6) (2) (3) (3) [14] Please turn over Downloaded from hlayiso.com

TOAKEKNZO)T MARKING GUIDELINE ENGINE! 6G SCIENCE NS QUESTION 7? 74 The heat developed when an electric current passes through a conductor is proportional to the resistance , the time and the square of the current. (3) 72 7.24 Rr = Rpor + Reer =(Art hy +3405" = 1,33 + 3,5” = 4,63 OY (4) 722 lp =V/Rr = 6/4,83¥" = 1,242 AY (2) Ix Rp V2 = br ® Ra 1,242 * 1,33” = 1,242 x 3 1,652 Vv = 3,726 Vv (4) 7.2.3 Vi nun 724 — Ig=VplRs = 1,652/4v 7 = 0,413 AY (3) [16] QUESTION 8 8.1 Metals « Ductile « Malleable * Good conductors of heat and electricity {Any 2x 1) Non-metals e Brittle » Poor conductors of heat and electricity (2% 2) (4) 8.2 e High temperature e High friction ® Heavy shocks {Any 2 1) (2) [6] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA : ‘Te60(E\(M30)T REQUIREMENTS 1 information Please turn over Copwiantresene? ~— Downloaded from hlayiso.com

(18070413) 20 TEGO(E}M30)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION Answer ALL the questions. 2. Read ALL.the questions carefully. 3. Nu in this question pa 4. AL 4.1 4.2 4.3 5. Drawing instrume! draw or diagrams. 6. All drawings or diagri 7. The constant values, a ched in rmation sheet, must be used wherever possiblé: 8. Keep subsections of questions together. 9. Rule off on completion of each question. 10 Use g = 9,8 m/s? 11. Answers must be rounded off to THREE decimal places. 12, Write neatly and legibly. Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) 3 TE6O(E)(M30)T QUESTION 1: MOTION, POWER AND ENERGY 1.1 Give the definition of a force. (2) 1.2 Give the definition of work done and power. (2) 1.3 An object has a mass of 2 Mg and rests on an incline with an angle of 10°. The object is to be released to slide 18 metres down the inclined plane to the bottom, neglecting all friction. |». Determine the following: — 41.3.1 The potential energy be f (3) 1.3.2 (1) 1.3.3 (2) 1.4 ontal road from 1 N is applied (4) [14] QUESTION 2: MOMENT: 2.4 Define the term ‘m (2) 2.2 A beam ABCDE wi supports atB and E. T : D. Concentrated or point loads at points A and C and another one of 4kN at point D. The lengths of the various portions are as follows: AB=2m, BC =2m, CD=4mandDE=3m. Use the information given above to answer the following questions: 2.2.1 Draw the beam described above. (2) 2.2.2 Calculate the reactional forces at B and E. (6) 2.2.3 Draw the shear force diagram with ALL the main values indicated on the diagrams. (3) [13] Copyright reserved Downloaded from hlayiso.com"**” turn over

(15070443) -4- T66O(E)(M30)T QUESTION 3: FORCES 3.1 Explain the term vector quantity. (2) 3.2 Complete the following sentence by filling in the missing word or words. Write only the word or words next to the question number (3.2.1- 3.2.2) in the ANSWER BOOK. A direct force is a force which acts (3.2.1 ...) and (3.2.2 ...) onto an object. (2) 3.3 The system in FIGURE 4 below is in a state of equilibrium. Determine analytically the magnitude of the unknown forces GP and T. (6) 3.4 Analytically determine members AD and DC of th RE 3 below. 4kN D | FIGURE 3 A | 2KN (4) 14] Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) 5 TEGO{E)(M30}T QUESTION 4: FRICTION 41 Give FOUR practical examples of kinetic friction. 4.2 A steel ball with a mass of 450 kg rests on an inclined plane of 12° to the horizontal. The force required to overcome the friction is 450 N. Calculate the following: 4.2.1 The weight component. perpendicular to the inclined plane 4.2.2 The coefficient of frictic (2 * 2) 4.3 An object with a weight of 4000 N of friction is 0,30. A force (P} with C nt velocity. a-horizontal plane. The coefficient the horizontal pulls the tude of force P. QUESTION 5 5.1 5.2 5.3 Determine the final length of the b: 5.4 A steam plant produced steam in a boiler with a pressure of 1 400 kPa and a dryness factor of 0,9. This wet steam now flows to a super-heater where it is dried and superheated to a temperature of 210 °C. If the specific heat capacity of superheated steam is 2,5 kJ/kg. °C, determine the following: 5.4.1 The enthalpy of the dryness saturated steam 5.4.2 The required enthalpy to change 2 kg of water into superheated dry steam (4) (4) [12] (2) (4) (3) (1) (3) [13] Copyright reserved Down ] oaded fro m h layi so.com Please turn over

(15070413) -6- T660(E})(M30)T QUESTION 6: HYDRAULICS 6.1 Explain the following terms with reference to hydraulics: 6.1.1 Pressure at a point 6.4.2 Total pressure (2 * 2) (4) 6.2 The average pressure in a single-acting piston pump installation is 158 kPa during a delivery stroke. .The piston diameter of the pump is 86 mm and its stroke lengthis 120mm. =. Caiculate the following: 6.2.1 (3 x 2) (6) 6.3 (2) [12] QUESTION 7: ELECTRICI 71 A student is testing a lls, each with an EMF of 2 volts and an inte nnected in series. An ammeter with an internal. s is connected in series to the circuit, with resistor Ra = resistor R Q connected in parallet. Calculate the following: 7.1.4 The total internal resistance of the battery (1) 7.1.2 The ammeter reading (3) 7.2 In an electroplating laboratory a steel casting is to be plated with copper. An ammeter connected to the circuit registers a current of 8 A. If the electrochemical equivalent is 3,3 x 10% g/C, determine how long it will take to form a layer of 220 gram copper. (3) Copyright reserved Please turn over Downloaded from hlayiso.com

(15070413) “7? T660(E)(M30)T 7.3 An electric bulb is marked 220 V — 80 W. An ammeter in the circuit shows that the bulb uses a current of 0,3 A. The bulb burns for one hour. Calculate the following: 7.3.1 The electrical energy consumption (2) 7.3.2 The cost for the bulb to burn for one hour if the cost of electricity is 26 cents per kW/h. (3) 74 Determine the power rating: of a stove if it releases 180 kJ of heat energy in 5,5 minutes. : : (3) ] [15] QUESTION 8: CHEMISTRY 8.1 (2) 8.2 Name ONE conditi against corrosion (2) 83 (2) 8.4 (1) [7] TOTAL: 100 Copyightreseved Downloaded from hlayiso.com

(18070413) -1- T660(E)(M30)T ENGINEERING SCIENCE N3 FORMULA SHEET All the formulae needed are not necessarily included. Any applicable formula may also be used. W= Fs Mm) .U,LMy Uy =m .V,tmMy.V7 W=p.v D,=(D+0) WV P= > Pnatavet = hy + wi jg __ Uitse/Output 100% PelaPn T= Fr Inset/Input aD? F=ma \ pate Nr f=tan® Fy = wsin@ Fo = weos0 Freosa = Fi, £ Fg... £,=T,-T, Ty; . . — = fension ratio 2 n ‘Oz=m.c. At..tp =lotAt P=F,.v m.ww = O= miv ver.d.nune® _2 60 ~ t Wy = Fy s AL = Loa. At... Lp=Lo® AL AE, =m.g. dh 4A= Ay. B. At... Ap= Ag AA Lasav—-0 O=l’ Rut seule tar mal.zt veutat Ve _ Np _ Is. STFH=SVE "s Ns Jp M=F.1s Copyright reserved Please turn over Downloaded from hlayiso.com

(18070413) -2- 2 CH AL = SACHA Pags = Paras + Parr P= dSxgxh i i i =H +— Rear Ri Ry Rseg = Ry +. Ry V,-V, =- e(U, -U5) V=IxR T680(E)(M30)T Copyisntreseved ~~ Downloaded from hlayiso.com

(15070413) INFORMATION SHEET TE6O{E)(M30)T PHYSICAL CONSTANTS VALUES FISIESE KONSTANTE Atmospheric pressure 101,3 kPa Atmosferiese druk Density of copper 8900 kgim> Digtheid van koper Density of aluminium 2770 kgim” Digtheid van aluminium Density of gold 19 000 kgim> Digtheid van goud Density of alcohol (ethyl) “790 kg/m" Digtheid van alkohol (etiel) Density of mercury _ 43.600 kgim* Digtheid van kwik Density of platinum Digtheid van platina Density of water Density of mineral Density of air Electrochemica' Electrochemical & Gravitational accele! Heat vaiue of coal Heat value of anthracite Heat vaiue of petrol armtewaarde van petrol Heat value of hydrogen mtewaarde van waterstof Linear coefficient of expansion of co| Linéére uitsettingskoéffisiént van koper Linear coefficient of expansion of aluminiti Lineére uitsettingskoéffisiént van aluminium Linear coefficient of expansion of steel Lineére uitsettingskoéffisiént van staal Linear coefficient of expansion of lead 54 x 10°F C Lineére uitsettingskoéffisiént van lood Specific heat capacity of steam 2100 dikg °C Spesifieke warmtekapasiteit van stoom Specific heat capacity of water 4187 Jikg.°C Spesifieke warmtekapasiteit van water Specific heat capacity of aluminium 900 J/kg.°C Spesifieke warmtekapasiteit van aluminium Specific heat capacity of oil 2 000 J/kg.°C Spesifieke warmtekapasiteit van olie Specific heat capacity of steel 500 J/kg.°C Spesifieke warmtekapasiteit van staal Specific heat capacity of copper 390 J/kg.°C Spesifieke warmtekapasiteit van koper Copyright reserved Downloaded from hlayiso.com

“ ent: Higher Education and Training REPUBLIC OF SOUTH AFRICA D MARKING GU NATIONAL CERTIFICATE APRIL EXAMINATION ENGINEERING SCIENCE N3 30 MARCH 2017 This marking guideline consists of 9 pages. 4 Chief Marker: J. “7 [ere ACHE Signature: han i Chief Marker: DYVY Wr ona Signature: (Lili iw all —to5 ee Internal Moderator: _/ 1G Sr LMANE _ Signature: —— Signature: Ve, DMCM: pi FAR HELE @). Lf £ ee MCN: M NTAMBOD X He Signature: \ bo Copyright reserved umatust: Ns LAKHA 4 ) Signature Downloaded from hlayiso.corff*°""

MARKING GUIDELINE Z ENGINEERING SCIENCE N3 QUESTION 4: MOTION, POWER AND ENERGY TEGCIE}MSO)T 14 A force is that influence which changes (or tends to change) the state of rest of a body or, if the body is already moving, a force is that influence which brings the body to rest (or tends to bring it to rest), or changes the direction of the body. (2) Hig the force appli n object and the object moves in the direction of the applied forse of Hoan be wark done per second. 1.2 Work done is the ability to do work and power is the rate at which work is done. (2) 13 13.4 h= 18 sin 10° =3.126m vv 1m value ofh PE = mgh = 2000x9.8x3126 ¥ 1m sub = 61269.6 J v 1 msolution = 61.269 KJ (3) 4.3.2 KE at bottom = PE at top y 61269.6 J= 61269.6J 1m solution (t) 1.3.3 KE=0.5xmxv" v= VKEx2/m : 4msub = 461269.6 x 2/2000 y = 7.827 mis v 1m solution (2) 1.4 Data given m = 200 000 g = 200 kg u=Om/s v=265 ms t= 50 seconds Fu = 900N a=v-u=25-0 v1 mark sub t 50 a=0,5 mis? ¥ 1 mark solution (2) Fr-FrtFy =ma+Fu = 200 x 0,5 + 900 v1 mark sub Fr=1000N ¥ — 1 mark solution (2) [14] Copyright reserved Piease turn over Downloaded from hlayiso.com

8 GUIDELINE “3 PSGGEL MM SOFT ENGINEERING SCIENCE N& QUESTION 2: MOMERTS 2.4 ‘Moment of a force’ is the force multiplied by the perpendicular distance or (2) 2.2 2.2.1 3kN i; 3 kN _4aky | | vv 5 5 kN/m | gov roa a . Soe go _ - A 8 (2) 2.2.2 Taking moments about B 3(2) + E(Q) = 3(2) + 20(4) + 4(6) “ SE = 104 v E= 11,556 kN Taking moments about E B(Q) = 3(11) + 3(7) + 20(5) + 4(3) ¥ 9B =166 v B= 18,444kN ¥ (6) 2.2.3 15,444m | rs : | 3 12,444" SFKN) ; ' i) O° larger @) [13] QUESTION 3: FORCES 3.1 A vector quantity is a quantity which possesses magnitude and direction. (2) 3.2 3.2.1 Directly 3.2.2 Perpendicularly (2 * 1) (2) Copyrightreseved = Downloaded from hlayiso.conie*" >

766 MEMO) MASRKING GUIDELINE 3.3 P gin 45 soon A 45° P Cos 45° For Equilibrium conditions: HC = 0 500 ~ P cos 45°70 v P = 500/cos45° v P = 707.107N ¥ VC =0 T—Psin 45°=0v T = 707.107 sin 45°v T= 500N ¥ OR Tan 45° = T/500 T = 500N Concession due to mistake on question (GP instead of P): e Candidates will be credited with 3 marks at minimum i the whole anewer is wrong. e ifthe student uses the wrong answer from the first part of the question to calculate T, full marks will awarded, (6) 3.4 CD = 2/sin 30°v = 4kN(strut or pull) ~ AD = 2/tan 30°V = 3.464(tie or push) KN ¥ (4) [14] Please turn over Copyright d i “pes Downloaded from hlayiso.com

MARKONGGUIDELINE “Be TEGO ENGINEERING SCIENCE N3 QUESTION 4: FRICTION 4.4 e A car's tyres on the read surface causing friction so that the car can pull away and move e Brake pads of a car for braking e Machining such as drilling - machining is only possible due to friction e Clutch to change gears (4) NR “ Vee Wxsin12° 72° 4.2.1 The weight component perpendicular with the plane =W Cos @ = 450 x 9,8 x Cos 12°v = 4313,631 NW 4.2.2 The coefficient of friction w= F pN = 450/w cos 12°v = 0.104 ¥ (4) 43 W=4000N;=0,30;0= 15° P=? P Sint5 Fu P Cos 15 q Copyrigntreseved = Downloaded from hlayiso.comie «7

f ENGINEERING SCIENCE NB OW Pints 0,30 = nee Py wo 4 000- Psinl5 F, = 0,3(4 000- Psin15) ......... Oy Peosi$ = Fu «. Pcos15 = 0,3(4 000 - Psin 15) Px 0,966 = 1 200 -0,078P v Px 0,966 + 0,078? = 1200 1200 v ~ (0,966 + 0,078) =1149,495N 9 ¥ (4) [12] QUESTION 5: HEAT 5.1 Energy can neither be created nor destroyed, but only converted from one form to another. (2) 5.2 My xox At=M, xc x At 5 x 4 187 x (t2—17)¥ = 21 * 500 « (89 — th)” 20 935 x (t2~ 17} = 10 500 * (89 —t,) 20 935t, — 355 895 = 934 500 - 10 500te” 31 435te = 1 290 295 te = 41,049 °Cv (4) 5.3 f=5,5m t,= 293 K f= 423 K Al = |,aAt = 5,5 x 23 x 10% x (423 - 293)v = 0,0164 mv Lr=Lo+Al = 5,5 +0,0164 = 5.5164 mv (3) Copyright reserved Downloaded from hlayiso.com

MARKIMGGUID ELIE “Fe ENGINEERING SCIENCE NS 5A 641 Awe) = hf + hig = 830 + (1 958) = 2788 kJ/kg 54.2 tigu & hg + Cp(iss-ts) hea = 2 788 + 2,5 x (210 - 195)” hey = 2 825.5 x 2 = 5 651 kJ QUESTION 6: HYDRAULICS 1) (3) [43] 6.4 6.1.1 Pressure at a point in a liquid is the force exerted by the liquid due to its weight per unit area, acting downwards. 6.1.2 Total pressure on an area means the force or weight acting over the whole area. 6.2 6.2.1 P =158 kPa d = 86 mm = 0,086 m L=120 mm=0,12 m VaVxL md* - rv 4 x 2 =O. 0,12 = 6,971x10 4m? 6.2.2 W=PxAxL = 158% (0,086)? x 0,12 =ll0s 6.2.3 t 0,110 a “30 x 60 6111 x10 ~S kw = 0,061 Gopyrightreseved == Downloaded from hlayiso. (2*2) (4) (2) (2) (2) lease turn over coni

MAR RINGGUINELINE G SCIENCE NG 6.3 Pneumatic equivalent is the pressure exerted and expressed as number of atmespheres. A pneumatic equivalent of 3 is equal to 3 * 101,3 kPa = 303,9 kPa. (2) 12) QUESTION 7: ELECTRICITY 7100 744 6x05 30 (1) 7.1.2 I= E R+r v _ 2 2,45 +3 vw = 2,202 A (3) 7.2 t= m IXZ 220 Y *'8x3,3x 10 = 83333 ,333 » = 23h,8min, 535 ¥ (3) v 7.3 7.3.1 V=220V P=80W I=0,34 t=] hour = 60x60 =3 600 Q=IVt = 0,3 x 220x3 600 = 237 600) OR 237,6 kJ (2) 7.3.2 Time = 1 hour unit = kWh _237,6xlh * a = 0,396 kW Ax 26 =1,716 cents v (3) Copyright reserved Please turn over Downloaded from hlayiso.com

MAR RINGGUIDELINE - ENGINEERING SCIENCE Né 74 Q=180000/ 1 = $min30s=5x60+30 = 330 s pal t - 180.000 330 v = 545,455 W (3) [15] QUESTION 8: CHEMISTRY 8.1 8.1.4 e Table salt e Sodium chloride {Any 1 x 1) (1) 8.1.2 e Table salt e Marble or calcium carbonate (Any 1 x 1) (1) 8.2 © Metals which are subject to friction e Aopen fires on or near the metal ¢ Very wet conditions (Any 2 x 1) (2) 8.3 Different ions in tap water can act as current carrier ions. Purified water is 100% pure and no ions exist to act as charge carriers OR Tap water has salty (2) minerals (ions) which contains minerals and conducts electricity, Purlfied water is 100% pure and no ions exist to act as charge carriers 84 Eight electrons (1) 7] TOTAL: 100 Copyright reserved Downloaded from hlayiso.comi**"" over

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higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA T520(E)(N17)T ER EXAMINATION NOVE ‘TIONAL CERTIFICATE 09:00-12:00 This question pap consists of 7 ages, 1 information sheet and1 formula sheet. Copyrightreseves DOwnloaded from hlayiso.comM p.ss0 wm over

(15070413) -2- TS20(E)(N17)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 10. 11. tution of the given’ 4.2 43 answer with the\eor The constant*values, as they appear on the attachment information sheet, must be used when. possible. Rule off on completion of each question. Drawing instruments must be used for all drawings/diagrams. ALL drawings/diagram must be fully labelled. Use g = 9,8 m/s” Answers must be rounded off to THREE decimals. Write neatly and legibly. Copyrightreseved Downloaded from hlayiso.com Please tum over

(45070413) -3- T520(E)(N17)T QUESTION 1: MOVEMENT 1.1 1.2 1.3 QUESTION 2: MOMENTS “- 2.1 Copyright reserved Calculate the final velocity in m/s of an object if it is moving in astraight line at a constant velocity of 60 km/h and then accelerates uniformlyat 2 m/s? for 4 seconds. A crane hoists a load of 2,2 tons by means of a chain. Thecrane hoisting drum has a rotational frequency of 60 r/min and a diameter of 750 mm. Calculate the following: 1.2.1 The torque requi ‘o.hoist the load 1.2.2 The rope sp the load in m/s 1.2.3 The work by the crane 1.2.4 uired for the The powel The efficiency of the’crane if the crane ruses 80 kW to drive the lifting hoist ith a mass of 50 kg § the barrelof the cannon of 10 tons fires a bull annon with a A beam ABCDE which rests horizontally is 12 m long and is supported at B and £. Support B is 3 m away from A and support E is at the aght-hand side of the beam. The beam carries point loads of 15 KN and 1,02tons at C and D respectively. The beam also carries a distributed load of 5 kKN/m for 5 meters starting at A and ending at C. It also carries a second dstributed load of 8 KN/m over the last 4 m between D and E. 2.1.1 Draw the beam, showing ALL critical values. 2.1.2 Calculate the magnitude of supports B and E. 2.1.3 Draw a shear-force diagram and fully label ALL critical values. (1%) (12) (1%) (1%) (2) (4) (15] (4) (6) (3) [13] Downloaded from hlayiso.com 5,356 turn over

(15070413) -4- T520(EKN17)T QUESTION 3: FORCES 3.1 Explain the difference between a resultant force and an equilibrant force. (3) 3.2 Determine the magnitudes of the forces P and Q shown in FIGURE 2. The system is in equilibrium on the same plane. 500 N 600 N 4 ——_¥.------------------- FIGURE 2 (8) , 3.3 a structure af foré Analytically‘determine or calculate the magnitude and nature of the members AB and AC. 150 kN 150 kN FIGURE 3 (4) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

(15070413) -5- 7520(E)(N17)T QUESTION 4: FRICTION 4.1 4.2 4.3 4.4 QUESTION 5: HEAT 5.4 5.2 5.3 Copyright reserved List ONE advantage of friction in practice. Name TWO disadvantages of friction in practice. Give the definition of angle of rest. A vehicle with a weight of 6 500 N is kept in equilibrium on an incline by means of a pulling force which is at an angle of 25° to the incline. The incline forms an angle of 35° to the horizontal plane. The coefficient of friction between the surfaces in ¢ 0,3. Determine the minimui pporting force to keep the object in equilibrium. heated from 293 K to is 26 MJ/kg and the (3 x 2) An object with a total mass of 1 500 g and a temperature of 81 °C is placed in 2 200 m? of water with a temperature of 20 °C. The final temperature is 25 °C. Determine the type of material used if the specific heat capacity of water is 4,2 kJ/kg °C. Use the steam tables to calculate the total enthalpy required for 1 kg wet steam for each of the following cases at the pressure of 1 400 kPa: 6.3.1 Wet steam with a dryness fraction of 0,94 §.3.2 Dry saturated steam 5.3.3 Total heat of superheated steam if the temperature of the superheated steam is increased to 240 °C and an average specific heat capacity of 2,85 kJ/kg °C. Use the boiler pressure as 1 400 kPa. Downloaded from hlayiso.com (1) (2) (2) (6) [41] (6) (3) (2) (1) (2) [14] Please turn over

(15070413) 6- TS20(EXN17)T QUESTION 6: HYDRAULICS 6.1 The efficiency of a pump is 80%. The pump delivers a quantity of 15 m* water per hour. Calculate the following: 6.1.1 The energy required if the water is pumped to a height of 8 m 6.1.2 The energy required if the water is pumped at a pressure of 600 kPa (2*3) (8) 6.2 A hydraulic jack is req to lift an object. The diameter of the ram is 80 mm and that of the plungef is 20 mm. The mechanical advantage is 16 and the stroke length of the plunger is 25 mm. The effort requi jack to lift a iss of 1 500 kg if the mechanical efficiency is 80% (4) The numb Sr of strokes required to lift th 51,5 mm (2) : [12] QUESTION 7: EL! 7.1 each with an EMF of 1,5 volts and internal resistance of 0,15" _are connected in series. This battery is then connected to two connected in parallel having resistors of 6 ohms and 3 ohms. Calculate the following: 7.1.1 The total resistance of the circuit (4) 7.1.2 The total current flow (2) 7.1.3 The internal potential drop in the battery (2) Copyright reserved Downloaded from hlayiso.com Please turn over

(15070413) 7.2 The following appliances are used for a given period of time per day: e Ten 60 Watt lamps, 4 hours per day e Five 100 Watt lamps, 6 hours per day e One 2 000 Watt stove, 1 hour per day s Two computers using 75 Watt, 3 hours per day Electricity costs R 1,63 per kWh. Calculate the electricity cost for SEVEN days. in a laboratory using an electric current of one for 2 hours through a solution of copper de. The electrochemical equivalent of copper is 7.3 A learner conducted an e 100 A. The experime sulphate using a copp 3,204 x 10% g/C, ‘ copper deposited Calculate the mass oft describe the function of el olysis. TOTAL: Downloaded from hlayiso.com Copyright reserved 7. T520(E)(N17)T (4) (2) [14] [6] 100

(15070413) T520(E)(N17)T ENGINEERING SCIENCE N3 FORMULA SHEETS Ail the formulae needed are not necessarily included. Any applicable formula may also be used. We=F.s mu, Emu, =m,.V, tm, = * D,=(D+0 Output n= ee Meas wet = h, + wh yy dnput , , Fema P=2aT nT =P ar F, cos0, +.....+ F, cos@, =F, sind, +.......+ F, sind, R=JDFy + ZF,” f, . . ~—— = fension ratio 2 By P=F,» tand == , xX N veadn n=O Q=mcAl...d; = Wt At W, =F 8 m.w.w = Q = mhy AE, =m.g.Ah p=2 AE, =v AL = 1, @.At..L, = ly + AL Qa? Rt AA = Ay. B.At....A, = 4, + AA m=lzt 2as=v -u? V, N i Pa Fe seut+tat Vy, Ny Ip 2 veutat Tt F=DlF M=F.1 s =CWM =x ACWM Pras = Pai + Paver P=dxgexh Peet yt Rowe = R, +..-R, Rose R, R, Vi ~¥, =-eU, -U,) V=IxR Copyright reserved Downloaded from hlayiso.com

(15070413) INFORMATION SHEET PHYSICAL CONSTANTS QUANTITY CONSTANTS Atmospheric pressure 101,3 kPa Density of copper 8 900 kg/m? Density of aluminium 2.770 kg/m* Density of gold 19 000 kg/m* Density of alcohol (ethyl) 790 kg/m? Density of mercury 13 600 kg/m? Density of platinum 21 500 kg/m® Density of water 1 000 kg/m*® Density of mineral oil 920 kg/m? Density of air 1,05 kg/m? Electrochemical equivalent of silvi 4,118 mg/C Electrochemical equivalent of coppe: 0,329 mg/C m/s? Heat value © Heat value Linear coefficient ofe 3x 10°7C Linear coefficient of exp: 12 x 10°/'C Linear coefficient of expansion of lead 54 « 10°7C Specific heat capacity of 2 100 J/kg. C Specific heat capacity of wa 4 187 J/kg. C Specific heat capacity of aluminitt 900 J/kg. C Specific heat capacity of oil 2 000 J/kg. C Specific heat capacity of steel 500 J/kg. C Specific heat capacity of copper 390 J/kg. C Copyright reserved Downloaded from hlayiso.com TS20(E)(N17)T

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higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE NATIONAL CERTIFICATE NOVEMBER EXAMINATION ENGINEERING SCIENCE N3 17 NOVEMBER 2016 17 This marking guideline consists of 9 pages. Downloaded from hlayiso.com Copyright reserved Please turn over

MARKING GUIDELINE -2- ENGINEERING SCIENCE N3 QUESTION 5 See marking guideline for allocation of marks QUESTION 1: MOVEMENT 1.41 u = 60/3,6 = 16,667 m/s¥ v=eutat v = 16,667 + (2).(4)¥ v = 24,667 mis¥ 1.2 m = 2,2 tons m = 2,2 x 1000 = 2200 kg N = 60 rpm Q = 750 mm = 0,75 m r =0,375 1.2.1 Torque = F xr T = 2200 * 9,8 x 0,375V T = 8085 Nm” 1.2.2 vemrxdxn 60 v= x 0,75 x 60 V 60 v = 2,357 mis ¥ 1.2.3 Work done = Force x 2 x 1 xn Wd = 2200 * 9,8 x21 x60V 60 Wd = 135483,04 J or 135,5 kJ ” 4.2.4 Power = Force x v Wd = 2200 x 9,8 x 2,357 ¥ Power = 50,817 kWY 1.2.5 n= Po x 100 Pin n= 50,7954 x 100 “ 60 1 =84,7%Y 1.3 My x uy + me * U2 = (My +M2) XV 10 000 x Vav¥ + 222,222 x OV = (50+ 10 0000) Ov V3 =-1,1 m/s or in the opposite direction’ T520(E)(N17)T 6 marks (3) (1%) (14) (1%) (1%) (2) (4) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE 3 T520(E)(N17)T ENGINEERING SCIENCE N3 QUESTION 2: MOMENTS 2.1 2.1.1 45 4,02 5 kNim 8 BYRYAYA B bakes . 3m L 2m , 3 . 4 be bi A B c D E (4) 2.1.2 Taking moments about B = Anti-clockwise moments = 2 Clockwise moments (E x 9) + (25 x 0,5) = (15 x 2) + (1020 x 9,8 x 5) + (32 x 7) QE + 12,5 = 30 + 49,98 +224 v R = 32,387 NY (3) Taking moments about E = Anti-clockwise moments = = Clockwise moments (32 x 2) + (1020 x 9,8 x 4) + (15 x 7) + (25 x 9,5) = (Bx 9)v 64 + 39,984 + 105 237,5 B = 49,6 KN (3) 2.1.3 = +345 = + 24,6 +96 2-04 =-16 = 32,4 (3) [13] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- ENGINEERING SCIENCE N3 QUESTION 3: FORCES 3.1 3.2 3.2 3.3 Copyright reserved The resultant of two or more forces is those single forces that can replace’ the two or more force and still have the same effect.v The equilibrant of a system of forces, on the same plane,v is that single force which will balance the system of forces.” XHC=0 600 =P x Cos50° + 500 x Sin25°v 600 — 500 Sin25° = P x Cos50° ¥ 600 — 211,309 = 0,643 x P v P = 388,691 P = 604,496 N ¥ 0,643 VC =0 500 x Cos25° = P x Sin 50°+ QV Q =- 604,496 x Sin50° + 500 x Cos 25°“ Q = - 463,071 + 453,154” Q=-9,917NY 309 kN 1450 kN 150 kN Tan 45° = AC/BC AC = BC tan 45° AC = 150 x tan 45° AC = 150 KN tie Cos 45° = BC/AB AB = BC/cos 45° AB = 212 KN strat T520(E\(N17)T (3) (4) (4) [15] Downloaded from hlayiso.com Pease tum over

MARKING GUIDELINE 5 TS20(E\(N17)T ENGINEERING SCIENCE N3 QUESTION 4: FRICTION 44 e Enable a vehicle to brake e Transferring of power by means of a clutch or belt drives e Can cause to drilling, or cutting of materials (Any 1 * 1) (4) 4.2 e Surfaces to wear away Drilling, cutting of tools getting blunt e Friction causes unnecessary heat (Any 2 x 1) (2) 43 The angle of rest is the biggest angle” that may be reached between an incline plane and the horizontal plane, before the body starts” moving freely down the incline plane. (2) 4.4 F Fcos25 = 0,3 6 500 cos35 6 500 cos35 \ F apptied = Fu + W x Sina F x Cose = p(W « Cosa +F x Sine) + W x Sina v F x Cos(25°) = 0,3(6 500 x Cos35° — F x Sin25°) +6 500 x Sin35°~ F x 0,906 = 0,3(5324,488+ 0,4226 x F) + 3728,2468 F x 0,906 = 1597,3464+ 0,12678 x F + 3535,539v F x 0,906- 0,12678 x F = 5325,592V 1,033 x F = §325,592 Copyright reserved P = 155,462 N ¥ P= 5,2 kN (6) (11) Downloaded from hlayiso.com ,,,.... inover

MARKING GUIDELINE -6- T520(E)(N17)T ENGINEERING SCIENCE N3 QUESTION 5: HEAT 51 5.1.4 At=t—ty At = 87-20 At = 67°C ¥ (2) 5.1.2 Q=mxcxAt = 200 x 4187 x (67) ¥ Q = 56,11 Mv (2) 5.1.3 Qout = m * HV m = 561058005” 26000000 m = 2,158 kg or use 2,2 kg¥ (2) 5.2 Q loss = Q gain mxcxAt=mxcx At 1,5 x cx (B1—25) = 2,2 x 4187 x (25~ 20) ¥ 15xc¢x56=2,2 x 4187x5 c=22x4187x5 v¥ 1,5 x 56 C = 548,3 J/kg.°C material is steel” (3) 5.3 5.3.4 m=1kg Pr=1400kPa tsu = 240 °C ts = 195°C hf = 830 kJ/kg fg = 1 958 kd/kg and hg = 2 788 kJ/k horey = [hf + x hg] » m hywet) = [830 + 0,04 x (1958)]¥ hwet) = 2 670,52 kd ~ (2) 5.3.2 hidryy = hg = 2788 kJ (~%) (1) 5.3.3 hss = hg +Css(tsu — ts) 2 788 + 2,85 (240 - 195) ¥ 2 788 + 128,25 hyet= 2.916,25 kl ¥ (2) [14] Copyright reseved Downloaded from hlayiso.com Pease tum over

MARKING GUIDELINE -7- ENGINEERING SCIENCE N3 QUESTION 6: HYDRAULICS 6.1 6.1.1 Pr=pxg*h Pr = 15 000 x 9,8 x 8” P= 1176 kPa(v’) n= Wdou * 100 Wd in Wdin = 1176 x 100 80 Energy = 1470 kJ ~ 6.1.2 Power = Prx V = (600 x 10°) x 15¥ P = 9000 MJY Power = 9 000 x 100 80 Energy = 11,25 MW v 6.2 6.2.4 fed? xW D? = (20) x 1500 x 9.8” (80) f= 918,75 NY Fi, = 918,75 x 100 80 Fi, = 1148,4375N MA = Load Effort Effort=Load = 1148,4375 MA 16 Effort = 71,77 Nv 6.2.2 d? x hx no of strokes = D°H (20) x 25 x no = (80) x 51,5” No of strokes = 32,96 use 33” Copyright reserved Downloaded from hlayiso. T520(E)(N17)T (3) (3) (4) (2) [12] com Please turn over

MARKING GUIDELINE ENGINEERING SCIENCE N3 QUESTION 7: ELECTRICITY 7.1 7.1.1 Roa RR, °R, +R, _ 3x6 643 =2Q Rp = Rp + Rin R, =2+0,9 R, =2,9 Q 7.1.2 V =6x1,5 =9 Volts pak R _ ov 2,9 Q = 3103 A 7.13 V = IR, =3,103 Ax 0,9 = 2,793 Volts 7.2 10x 60x 4 _ 24 kWh 1000 5x100x6 =3 MWh 1000 2 000 x1 =? kWh 1000 2x 753 =0,45 kWh 1 000 2,443.42 +0,45 = 7,85 kWh 7,85 kWh x Tdays 54,95 kWh x R1,63 R89,57 Copyright reserved 1 M substitution 1 MR parallel 1M sub 1M answer 4 Msub 4M answer 2 M ten lamb % M five lamps % M Stove Y% M Computers 1M 7,85 kWh 1 M R89,57 T520(E)(N17)T Downloaded from hlayiso.com Pesse tum over (4) (2) (2) (3)

MARKING GUIDELINE -O- T520(E)(N17)T ENGINEERING SCIENCE N3 7.3 1=100A t= 2 hours = 2 x 60 x 60 = 7200 sec z= 3,294 x 10“g/c Calculate the mass =| «tz m = 100 x 7200 x 3,294 x 10% v m = 237,168 g ¥ QUESTION 8: CHEMISTRY Battery |] pt Sten ly [solated copperwire GALVANOMETER Anode = Cathode Platina plates Electrolyte fs tass container TWO marks for sketch and labels The electrolytic process requires that an electrolyte,“ an ionized solution or molten metallic salt, complete an electric circuit between two electrodes. When the electrodes are connected to a source of direct current one, called ihe cathode, becomes negatively (-) charged while the other, called the anode, becomes positively (+) charged.v The positive ions in the electrolyte will move toward the cathode and the negatively charged ions toward the anode. This migration of ions through the electrolyte constitutes the electric current in that part of the circuit.” The migration of electrons into the anode, through the wiring and an electric generator, and then back to the cathode, constitutes the current in the external circuit.” TOTAL: Downloaded from hlayiso.com Copyright reserved (2) [14] [6] 100

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higher education & training Department: Higher Education ard Training REPUBLIC OF SOUTH AFRICA This question paper consists of 10 information sheet and1 formula sheet. Downloaded from hlayiso.com Copyright reserved Please turn over

(15070413) -2- T580(E)(M30)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMA 10. Answer all the questions. All calculations shi onsist of at least THREE steps: Rule off on completi Drawing _ instruments all drawings/diagrams. All drawings/diagrams must be Use g = 9,8 mis’. Answers must be rounded off to THREE decimal places. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Pease tum over

(15070413) -3- T580(EXM30)T QUESTION 1: MOVEMENT 14 A car travelling at 20 m.s” applies its brakes and stops after 50m. Determine the following: 1.44 The average acceleration 4.1.2 The time taken to stop (2 * 2) (4) 1.2 (3) (1) (1) i vehicle with a mass of 2,5 tons in 25 and reaches a speed o 1.3 ccel mivatier 3 mi utes. Caicula 1.3.4 (2) 1.3.2 (2) 1.3.3 (3) [16] Downloaded from hlayiso.com Copyright reserved Please turn over

(15070413) -4- T580(E)(M30)T QUESTION 2: MOMENTS 2.1 State the definition of a force. (2) 2.2 A uniform beam is 18 m long and rests horizontally on two supports. The one support is at the left end and the other support is 4 m from the right end. The beam carries concentrated loads of 200 N and 100 N 5m = and 10 m from the left end respectively. The beam also carries a uniformly distributed load of 15 N/m over the first 6 m from the right end. actions at the supports and test your answers. (5) (4) 2.2.1 Calculate FIGURE 2 (2) [13] Copyright reserved Downloaded from hlayiso.com Please turn over

(15070413) 5 T580(E)(M30)T QUESTION 3: FORCES 3.1 State the definition of equilibrant of forces. (2) 3.2 Solve analytically the magnitude and direction of the resultartin FIGURE 3 below. 300 N (4) (5) . Copyright reserved Downloaded from hlayiso.com.,,.... turn over

(15070413) ~6- T580(E)(M30)T 3.4 Calculate the nature and magnitude of forces with reference to the components AB and BC in the roof truss in FIGURE 5 below. 500 N A B 60° _ 30° [oe 350 N 150 N (4) [15} (2) (2) 4.3 The weight of an object in FIGI below is 2 500 N. The object is placed on an inclined plane that for angle of 20° to the horizontal. The coefficient of friction is 0,4 for the: urface. FIGURE 6 Copyright resewed Downloaded from hlayiso.com Pease wm over

(15070413) -T- TS80(E)(M30)T Calculate the following: 4.3.4 The pushing force P required to push the object upwards against the slope (6) 4.3.2 The angle in degrees that the plane must be raised so that the body will slide down by itself (4) [11] QUESTION 5: HEAT 5.1 (2) 5.2 g °C and that of oll 1,5 kJ/kg ° (4) 5.3 overhead crane is used to carry a mould in workshop. The of the overhead crane has a length of 49 mat a temperaure of 18 °C. moulding process the tempera lre‘ofsthe beam of the overhead xpected to rise {9.100 °C in the dry. n which is 25x 10°C, (1) (2) §.3.3 The expected final lengt of the beam in metres (2) Downloaded from hlayiso.com Copyright reserved Please turn over

(15070413) 8- T580(E)M30)T 5.4 4 Water Water &Steam Steam le <<. —— P< Temperature in °C ¥v FIGURE 7 above shows a waco missifg descriptions and I plete the process in FIGURE 7 by ling the “symbol next to the question (4) [15] QUESTION 6: HYDRAULI 6.4 Explain the following terminology in hydrauli 6.1.1 Hydraulics (2) 6.1.2 Hydrostatics (1) 6.1.3 Fluidity (2) 6.2 A single- -action water pump with a piston diameter of 120 mm has to deliver 4,4 « 107 m? water per stroke. The average force exerted on the piston is 2,4 KN per working stroke. Calculate the following: 6.2.1 The pressure of the fluid during the working stroke (3) 6.2.2 The length of the stroke (2) 6.2.3 The work done per stroke length 3) Copyright reserved Downloaded from hlayiso.com Please turn over

(15070413) -9- T580(E)(M30)T QUESTION 7: ELECTRICITY 7.1 7.2 7.3 Copyright reserved Explain the concept power factor, and distinguish between a high power factor and a low power factor. (2) An electrical circuit as shown in FIGURE 8 below consists of 4 cells connecting in series, each having an EMF of 4 volts and a internal resistance of 0,15 ohms per cell, connected in series. R,F82Q Caiculate the follow 7.2.1 The total resista 7.2.2 The voltage across A (2 x 3) (6) A single-phase transformer has a supply voltage of 220 V. The turn ratio of the primary to the secondary is 12 : 1. The secondary current at full load is 20 A and the secondary coil has 50 windings. Calculate the following: 7.3.1 The secondary voltage 7.3.2 The primary current (2 x 2) (4) [12] Downloaded from hlayiso.com Please turn over

(15070413) -10- 7580(E)(M30)T QUESTION 8: CHEMISTRY 8.1 8.2 8.3 8.4 Name TWO precautions to keep moisture and water away from metals which can corrode. Name the chemical composition of the following compounds: 8.2.1 Limestone (marble) 8.2.2 Caustic soda (2* 1) Name a non-metal: conductor of electricity. Name ONE prop TOTAL: Copyrightreseves Downloaded from hlayiso.com (2) (2) (1) (1) [6] 100

(15070413) T580(EMM30)T FORMULA SHEET ENGINEERING SCIENCE N3 All the formulae needed are not necessarily included. Any applicable formula may also be used. We=Fs My ty Ems dy = mV, tinyv, ‘vy P= x D,=(D+t) itset / Output = Ciel! Cuuput Preciwer = hy + XR yy Inset / Input Fema Palain.P=Fr A PL tan b = —— xX Q=mcAt.t, stot At mw.w=QO=miav p=Z t AE, = Sim AL = Ly.a.At....L, =L,y + AL O=l’ Rt AA= A,.B.At...A, = Ay + AA malt 2as=v —u* i Vo Nop fs. s=utttat? Vs Ns P veutat StF=dIF M=F.1 s = CWM => ACWM Paes = Pry + Pur P=6xgexh 1 =14 sees gl Ryeg RR +R, Roar 2 R, ¥,~V, =-eU, -U,) V=IxR Downloaded from hlayiso.com Copyright reserved Please turn over

(15070413) INFORMATION SHEET PHYSICAL CONSTANTS 1580(E)(M30)T QUANTITY CONSTANTS HOEVEELHEID KONSTANTE Atmospheric pressure 101,3 kPa Atmosferiese druk Density of copper 8 900 kg/m? Digtheid van koper Density of aluminium 2770 kgim® | Digtheid van aluminium Density of gold 19 000 kg/m? | Digtheid van goud Density of alcohol (ethyl) 790 kg/m? Digtheid van alkohol (etiel) Density of mercury Digtheid van kwik Density of platinum 13 600 kg/m* Digtheid van platinum Density of water Digtheid van water Density of mineral oil Digtheid van minerale olie Density of air Digtheid van lug Electrochemical equivalent o silver 1,118 mg/C Elektrochemiese ekwivalent van silwer Electrochemical Quivalent of copper chemiese ekwivalent van koper Gravitational acceleration 9,8 mis? versnelling Heat valué of coal 30 MJ/kg mtewaarde van steenkool Heat nthracite 35 MJ/kg Warmtewaardé van antrasiet Heat strol 45 MJ/kg Warmtewaarde\ van petrol Heat value Warmtewaarde’ van waterstof Linear coefficient copper ttingskoéffisiént van koper Linear coefficient of ex aluminium neére uitsettingskoéffisiént van minium Linear coefficient of expansio steel ére uitsettingskoéffisiént van staal Linear coefficient of expansion of lead eére uitsettingskoéffisiént van lood Specific heat capacity of steam Spesifieke warmtekapasiteit van stoom Specific heat capacity of water Spesifieke warmtekapasiteit van water Specific heat capacity of 900 Jikg. 6 Spesifieke warmtekapasiteit van aluminium aluminium Specific heat capacity of oil 2 000 J/kg.°C | Spesifieke warmtekapasiteit van olie Specific heat capacity of steel 500 Jikg.°C Spesifieke warmtekapasiteit van staal Specific heat capacity of copper 390 Jikg.°C Spesifieke warmtekapasiteit van koper Copyright reserved Downloaded from hlayiso.com

higher education & training Deparment: Higher Education and Training REPUBLIC GF SOUTH AFRICA MARKING GUIDELINE NATIONAL CERTIFICATE APRIL EXAMINATION ENGINEERING SCIENCE N3 30 MARCH 2016 This marking guideline consists of 11 pages. con Downloaded from hlayiso.com ‘copyright reserved

MARKING GUIDELINE -2- T580(E(M30)T ENGINEERING SCIENCE N3 NOTE: ALTERNATE CORRECT ANSWERS MUST BE CONSIDERED. QUESTION 1: MOVEMENT 14 1.1.4 V2 =u? 42as yan a= 2s __ (0)? ~ (20)? ¥ Correct substitution 2(50) . > v¥ Negative answer only =-4 m/s” (2) 1.1.2 veu+at Vou te a t = 0) = 20) ¥ Substitution 4 =i s ¥ Correct answer (2) 120 1.2.1 veuret u=Vv~ ot = (0) — (-9,8)(3) g=-9,8 mis = 29,4 m/s vu value v —u? s= gt 2 2 - 0) -@9,4)" v Substitution -9,8(2) =44) m ¥ Correct answer u=? (3) 1.2.2 s=441ix2 = 88,2 m ¥ Correct answer (1) 1.2.3 u=v=29,4 mis ¥ Correct answer (1) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T580(E)(M30)T ENGINEERING SCIENCE N3 1.3 v= 55 kin/h 25% 1 000 = 12,278 mis 3 600 t=3minx 60 =180 s u=Omis m= 2,5x 1000 =2 500 kg 2,5 tons h 1 to 25 4.3.1 vou a= t 15,277 —0 ae v ituti 180 (“) substitution = 0,085 m/s (“) Correct answer (2) 1.3.2 E,=—mv" ] =5@ 500)(15,277)" (”) substitution = 291 733,411 J (“) Correct answer (2) 1.3.3 S=ut+ Yat? = (0) (180) + % (0.085) (180) (Y) s value =1377m Ep = mgh (”) Substitution © = 2500 (9.8) (1377/25) (v) Correct answer = 13494605 (3) [16] Downloaded from hlayiso.com Copyright reserved Please turn over

MARKING GUIDELINE -4- TS80(E}(M30)T ENGINEERING SCIENCE N3. QUESTION 2: MOMENTS 2.1 A force is that influence which changes or tends to change the state of rest of a body.(~) If the body is already moving, a force is that influence which brings the body to rest or changes the direction of the body.(v’) (2) 2.2 2.21 Taking moments about L: z Anti-clockwise Moments = = Clockwise Moments (R x 14) = (200 x 5) + (90 x 15) + (100 x 10)(“) E = 239,286 N(“’) (2) Taking moment about R: = Anti-clockwise moments = £ Clockwise Moments (100 x 4) + (200 x 9) = (Lx 14) + (90 x 1)(”) L = 150,714N (“) (2) Test: Upward forces = Downward forces 239,286 +150,714kKN= 390kN (“) (1) 2.2.2 150,714 (v) +60N ”“) (“) 0 0 ov -49,286 “ -149,286 N 179,286 N (4) 2.3 CosexFx5m=1,2x120+200x3,1m Cos30 x Fx 5 =144 +620 (v) F = 764 Cos30 x 5 F = 176,443 N (v) (2) [13] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE ENGINEERING SCIENCE N3 QUESTION 3: FORCES 3.1 The equilibrant of a system of forces in the same plane(~’) is thal single force which will balance the system of forces.(v) 3.2 HC = -150 + 216.506 + 141.421 = 207.927N Vv VC = 300 +125 — 141.42 = 283.579 N ¥ R = V (207.927)? + (283.579) = ¥ 123650.686 = 351.64N ¥ Tan 8 = 283.579 207.927 = 1.36 @ =Tan ‘1.36 0 N = 53.67° =E 53.67°N ¥ 3.3 P Psin35° PCos38° QCos40 > Q asin4o° Copyright reserved Downloaded from hlayiso.com T580(E)(M30)T Please turn over (2) (4)

MARKING GUIDELINE T580(E)(M30)T ENGINEERING SCIENCE N3 Ve=DtF=ETF P Sin35 =O Sin 40 P= Q Sin40 Sin35 P=11207 O veces (I) 2H PCos35+ QO Sin 40 = 560 (1,1207Q) Cos 35 + Q Sin 40 = 560 0,9180 + 0,766.0 = 560 O = 332,537 N Pe 295,82 Sin 40 Sin3s = 372,663 N 3.4 60° (“) Substitution (“) Equation (1) (”) Substitution (“) Value of Q (“) Value of P (5) ABSin60° 8s, ABCos60° 350 N Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -?- T580{E)(M30)T ENGINEERING SCIENCE N3 Slr=elter ead AB Sin 60" =350 (“) Equation _ 350 Sin 60° = 404,145 kN (7) AB Value LeKFaLoFr AB Cos 60 = BC 404,145 Cos 60 = BC BC = 202,225 kN (“) BC Value Member | Magnitude | Nature AB 404,145 Strut (“) Nature BC 202,07 Tie Strut a Tie — (4) [15] QUESTION 4: FRICTION 4.4 Static friction opposes the initial movement. The force required to just move a body is called the static friction force.(“’) Kinetic friction opposes the force causing the movement after the body is in motion.(”) (2) 4.2 e Braking pads of vehicles, trains, lifts and aircraft e Clutch engagement to transmit torque or power e Drilling, cutting, turning made possible by friction between cutting surface and material.(“) (Any 2 x 1) (2) 4.3 FIGURE 6 Downloaded from hlayiso.com Copyright reserved Please turn over

MARKING GUIDELINE 8- T580(E)(M30)T ENGINEERING SCIENCE N3 4.3.1 F applied = Fu + W x Sina P x Cose = u(W x Cosa + P x Sine)+WxSina = (v) P x Cos(30°) = 0,4(2 500 x Cos20° + P x Sin30°) + 2 500 x Sin20° P x 0,866 = 0,4(2500 x 0,9397 — P x 0,5) + &5,05 + (“) Px 0,866 = 939,7 —0,2 x P + 855,05 ) P x 0,866 - 0,2 x P = 939,7+ 8379,494 (7) 0,666 x P = 1794,75 (%) P=2694,82N (“) (6) 4.3.2 lan =u pe=tan' 0,4 = 21,8" Y (1) [41] QUESTION 5: HEAT 5.1 Energy cannot be created or destroyed(”’) therefore heat gained is equal to heat lost by an object.(“) (2) 5.2 m= 45x 0,75 = 33,75 t, =180 °C She sroe.y = S00 SIkg °C My =1 500 J/kg °C t, =20°C Heat lost by steel = Heat gained by oil mx she x (T2— 74) = mx she x (T3 — Ta) (“) 45 x 0,75 kg x 500 J/kg.°C x (180 — Tz) = 5 kg x 1 500 J/kg.°C x(T2 — 20) 16 875 x (180 — Tz) = 7 500 x (T2—-20) (“) 3 037 500 -— 16 875 x Tz) = 7 500 x T2— 150.000 7 500 T2 + 16 875 Tz = 3.037 500 + 150000 (~“) 24 375 Tz = 3 187 500 Tz = 130,769 °C ¥ 2 (“) (4) 5.3 7=19m t, =18 °C 1,=100°C a=5x10%C Copyrghtreseveed Downloaded from hlayiso.com presse ium over

MARKING GUIDELINE -9- T580(E)(M30)T ENGINEERING SCIENCE N3 5.3.1 At =100-18 on “) =82°C 5.3.2 At=toxax At AL= 19x 25x 10% x 82 (“) At = 0,03895 m or 38,95mm (“) 5.3.3 Le = Al+ fo Le = 0,03895 + 19 (v) Le = 19,03895 m (v) 5.4 5.4.1 he = liquid enthalpy 5.4.2 hf, = enthalpy of evaporation 5.4.3 hg = enthalpy of dry steam 5.4.4 hs = enthalpy of superheated steam (4x1) QUESTION 6: HYDRAULICS 6.1 6.1.1 Hydraulics refers to the pressure(“) and work done of fluids.(“) 6.1.2 Hydrostatics refers to fluids that are at rest. 6.1.3 Fluidity is a fluid or a gas which has no form(’) but takes up the form of a vessel in which it is captured.(v) 6.2 6.2.1 Volume =1,4x107 m3 F=2,4 kN 2 - 2 mt A=2(0,06)" or A= m1 (v) Area Vv “ae ti _ 2 400 (~’) Substitution ~ 0,0113 = 212,2 kPa (“) Answer Downloaded from hlayiso.com Copyright reserved (1) (2) (2) (4) 115] (2} (1) (2) (3) Please turn over

MARKING GUIDELINE -10- T580(E}(M30)T ENGINEERING SCIENCE N3 6.2.2 Stroke length = Volume rea 33 = bAxl0’ mn (7) Substitution 0,0113 m~ =0,1238 mor 123,8imm (“) Answer (2) 6.2.3 WD=Pxy =212 200x1,4x10° =297 Jistroke (2) [12] QUESTION 7: ELECTRICITY 7.1 Power factor ( (cosa) is the ratio of actual power or watts to volt amps.(“’) The smaller the phase angie the higher the efficiency and vice versa.(v’) (2) 7.2 7.21 Rp =Ri+Re+Rs RiRe+RiRs+ReR3 =8x5x5 8x5+8x5+5x5 (“) Substitution = 200 105 ~ Rp=1 19052 =1.905 Q Rt = 1.905 + 0.6+0.5+45+2 = 9.505 O 7.2.2 Emf=4x4=16V Rt=4x05=0.60 = 16 9.505 + 0.6 = 1.583 A V=IR = 1.583 x2 = 3.166 V (“) R7=9.505Q (3) (“) Substitution (“) Current (“) Volt drop (3) Copyright reseved Downloaded from hlayiso.com picase turn over

MARKING GUIDELINE -11- ENGINEERING SCIENCE N3 73° «734 Np _¥p Lg Ny Ve Tp yale N, 50% 220 yp» 1x220 600 12 = 18,33 ¥ 7.3.2 1, = Nskts OR Vex, N, ¥, - 50x 20 OR 18,33 x 20 600 220 = 1,667 A QUESTION 8: CHEMISTRY 8.1 By painting the metal By applying oil or grease By electroplating with zinc or tin By galvanising By using plastic film © 6@ © @ © 8.2 8.2.1 8.2.2 Caustic soda = NaOH 8.3 Carbon 8.4 e Low melting point e Soft e Good adhesion to other metals Copyright reserved Limestone (Marble) =CaCO3 (“) Substitution (“) Solution (“) Substitution (“) Solution T580(E)(M30)T (2) (2) [12] (Any 2 x 1) (2) (21) (2) (1) (Any 1 * 1) (1) [6] TOTAL: 100 Downloaded from hlayiso.com

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higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA ooove00000 ~T570(E)(N30)T NOVEMBER EXAMINATION ENGINEERING SCIENCE N- (15070413) 0 November 2015 (X-Paper) . 9:00-12:00, REQUIREMENTS: Properties of water and steam (BOE 173 This question paper consists of 7 pages, 1 information sheet and 1 formula sheet of 2 pages. Downloaded from hlayiso.com Copyright reserved Please turn over

(15070413) -2- T570(E)(N30)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION » : 1. Answer ALL the questions. 3. 4. The formula used or manipulation thereof Substitution of the given data in the formula _ The answer with the correct SI unit 5. 6. 7. 8. Drawing instruments must rawings/diagrams. ALL drawings/- diagrams must be fully labelled 9. Use g = 9,8 m/s”. 10. Answers must be rounded off to THREE decimal places. 11. Write neatly and legibly. Copyright reseved Downloaded from hlayiso.cOm piease tum over

(16070413) -3- T570{E}(N30)T QUESTION 1: MOTION, POWER AND ENERGY 4.1 4.2 1.3 1.4 Give the definition of a force. (2) If a force is applied to an object, it can have an effect on the objct. Name THREE effects of a force. (3x1) (3) An object with a mass of 1,9 tons and a resistance force of 280N travels at a speed of 120 km/h. Determine the following: : 1.3.1 The power required of the object at a constant speed squired to accelerate from rest with aceleration of 4.3.2 The power 4 4-mis?. (2x3) (6) A tractor with a mass of 4200 kg accele m res up an incline (2) (3) [16] QUESTION 2: MOMENTS 2.1 Give the definition of the La (2) 2.2 FIGURE 1 below shows a lig rizontal beam ABCDEF with a uniform cross-section, loaded as shown Calculate the reaction forces B and E and draw the shear-forcediagram using an appropriate scale. 30 kN 8 kim M/ ae ry FIGURE 1 (9) {11] Copyright reserved Downloaded from hlayiso.com Please turn over

(15070413) -4- TS70(E)N30)T QUESTION 3: FORCES 3.1 Give the definition of resultant of forces. 3.2 Caiculate the magnitude and direction of the resultant in FIGURE 3 below. 300 N 200N FIGURE 3 3.3 Calculate the turning momen slctum in FIGURE 2 below. FIGURE 2 QUESTION 4: FRICTION 41 Give FOUR applications of fri he engineering industry. (4x%) 4.2 The weight of an object is 4 and it rests on an incline of 30° to the horizontal. The coefficient of friction is 0,6 for the surface. Calculate the following: 4.2.1 The force required to pull the object upwards against the slope if the pulling force forms an angle of 20° with the incline. 4.2.2 The force required to push the object downwards along the slope if the pushing force forms an angle of 25° with the incline. (2) (11) (2) (5) (5) [12] Copyright reseved Downloaded from hlayiso.cOmM prase wm over

(15070413) 5 T570(E)(N30)T QUESTION 5: HEAT 5.1 5.2 5.3 5.4 QUESTION 6: HYDRAULICS 6.1 6.2 6.3 Copyright reserved Define the specific heat capacity of a substance. Determine the mass of silver when silver is heated up to a temperature of 120 °C and then placed in a container with 20 kg water at 50°C. The final temperature reached by the silver and water is 60 °C. Use the specific heat capacity of silver as 0,2 kJ/kg °C. A rectangular steel plate of a steam boiler has measurements of 1,5 m x 1 m. As the steam boiler is heated, its temperature increases from 20°C to 280 °C. The coefficient of linear expansion is,34 x 10° °C. Determine the final a Steam is developed si am. under pressure of 760 kPa and wet steam is 5.4.3 The required superhéating of dry steam (hes) : Distinguish between gauge p absolute pressure. A single-action suction pump has a diameter of 110 mm and adelivery stroke length of 75 mm and has to pump water to a vertical height of 30m. Calculate the following: 6.2.1 The volume of the water which is being pumped per delivery stroke 6.2.2 The mass of the water pumped per delivery stroke 6.2.3 The work done per stroke length (3 x 2) A hydraulic press has a plunger diameter of 75 mm. The force on the plunger is 50 kN while the ram diameter is 160 mm. Calculate the following: Downloaded from hlayiso.com (2) (3) (3) (2) (1) (4) [15] (2) (6) Please turn over

(16070413) -6- T570(E)N30)T 6.3.1 The mass in kg that can be raised (3) 6.3.2 The pressure of the fluid (2) [13] QUESTION 7: ELECTRICITY 7.1 An electrical circuit as shown in FIGURE 4 below consists of 3 cells connected in series, each having an EMF of 2 volts and internal resistance of 0,1 ohms per cell, conne ina series circuit. Calculate the following: 7.1.1 The total resistance (4) 7.1.2 The ammeter readin (2) 7.1.3 The voltage across AB (1) 7.1.4 The current through the 4 © resistor (2) 7.2 A single-phase transformer has a supply voltage of 220 V. The turn ratio primary: secondary is 10:1. The secondary current at full load is 2 A and the secondary coil has 28 windings. Calculate the following: 7.2.1 The secondary voltage 7.2.2 The primary current (2 x 2) (4) [13] Copyright reserved Downloaded from hlayiso.com Please turn over

(15070413) “7 T570(E)(N30)T QUESTION 8: CHEMISTRY 8.1 List SIX known metals. (4x6) (3) 8.2 Give TWO examples of reactions that cause corrosion. (2x1) (2) [5] TOTAL: 100 Copyright reseved Downloaded from hlayiso.com

(15070413) -1- T570(E)(N30)T ENGINEERING SCIENCE N3 FORMULA SHEET All the formulae needed are not necessarily included. Any applicable formula may also be used. W=Fs my .uptmy.up=M,.v, tM9.V> We=p.V D.<@+s Per Anaiwet = hy + xy t P2e2n.7Tn..T =F Fema ues Np a=0 % Fy = Fsin Fy = wsinO 2Fy = Fycos0; +... + F,cos0, Fo = weos0 «a ZF, = FysinO Freosa = Ft Fs ..a=0 Fo=1,-T, T; . . -—~ = tension ratio 2 P=F,.v ver dn. nad. “pu 60 1 W = Fy os AL =Lo.a. At... Lp=Lo+ AL dE, =m.g.Ah dA = Ay. B. At... Ap= Age A AEx =S.m.av? Das=v-u’ _ le Oz=l’ Rt saul+ sat m=L1.z.t veutat Vo Np _ Is EtFaEMF Vs Ns Ip dios . Copyright resevea Downloaded from hlayiso.com pizse tn over

(15070413) -2- 2 CWM = SACWM Paps = Para + Pyer P= 5x gxh 1 I I a Rear Ry R, Rep = Ry + Ry V,-Vy =~ e(U, -Up) V=IxR Copyright reserved Downloaded from hlayiso.com TS570(EMKN30)T

(15070413) INFORMATION SHEET T570(E)(N30)T PHYSICAL CONSTANTS QUANTITY CONSTANTS | HOEVEELHEID KONSTANTE Atmospheric pressure 101,3 kPa Atmosferiese druk Density of copper 8 900 kg/m? Digtheid van koper Density of aluminium “2.770 kgim® Digtheid van aluminium Density of gold 19 000-kgim* Digtheid van goud Density of alcohol (ethyl) 790 kg/m Digtheid van alcohol (etiel) Density of mercury Digtheid van kwik Density of platinum: -Digtheid van platina Density of water 920 kg/m? 1,05 kgim> 1,118 mg/C | Elektrochemiese ekwivalent van silwer 0,329 magic “{)Elektrochemiese ékwivalent van koper Gravitational acceleration 9,8 misz / ‘Sw : ekragversneling Heat value of coal 30Md/kg Warmtewaarde van steenkool Heat value of anthracite 35 Md/kg 7 armtewaarde van antrasiet Heat value of petro! 45 MJ/kg : | Warmtewaarde van petrol Heat value of hydrogen Warmtewaarde van waterstof Linear coefficient of expansion of copper Lineére uitsettingskeéffisiént van koper Linear coefficient of expansion of aluminium c | Lineére uitsettingskoéffisiént van aluminium Linear coefficient of expansion of steel © 12 * 10°C Line€ére uitsettingskeéffisiént van staal Linear coefficient of expansion of lead 54 x 10 PC Lineére uitsettingskoéffisiént van lood Specific heat capacity of steam 2 100 J/kg.°C Spesifieke warmtekapasiteit van stoom Specific heat capacity of water 4187 Jikg.°C Spesifieke warmtekapasiteit van water Specific heat capacity of aluminium 900 J/kg.°C | Spesifieke warmtekapasiteit van aluminium Specific heat capacity of oil 2000 J/kg.°C | Spesifieke warmtekapasiteit van olie Specific heat capacity of steel 500 J/kg.°C Spesifieke warmtekapasiteit van staal Specific heat capacity of copper 390 U/kg.°C Spesifieke warmtekapasiteit van koper Copyright reserved Downloaded from hlayiso.com

Depariment: Higher Education and Training REPUBLIC OF SOUTH AFRICA GUIDELINE NATIONAL CERTIFICATE NOVEMBER EXAMINATION ENGINEERING SCIENCE N3 05 NOVEMBER 2015 This marking guideline consists of 10 pages. Jet Week DM Mp Aime Abert Motberter hls mS yn prstetlfeod ofiafis Copyright reServed Pp onlooclon from hlayiso.combiese tum over

MARKING GUIDELINE ~2- TE7OCEHNGOVT ENGINEERING SCIENCE N3 QUESTION 4; MOTION, POWER AND ENERGY 14 A force is that influence which changes the state of rest of an object’ or body or, if the body is already moving, a force is that influence which brings the body to rest, or changes its direction of form.” (2) 4.2 e Causes a stationary object to move/acceleration ° Causes moving objects to come to rest v © Causes objecis to change direction v ° Cause objects to change their form/shape. (Any 3x 1) (3) 1.3 1.3.4 Power at a constant speed: Work done in one second = force x distance in one second P =F x S/sec = 280 N x (120 km x1 000 mv 3 600 sec = 280 N x 33,333 m/sv = 9 333,333 J/sec = 9 333,333 Watt’ (3) 1.3.2 Power to accelerate: Force required: F, = ma = 1900 kg x 4 m/s? =7 600 NV Fy = 280N Fr = Fut Fa | <= 280N+7600N Fy = 7 880 NY Therefore power PF xX Sisec P-= 7 880 N x 33,333 P = 262664J/sec P = 262,664 kwv (3) 1.4 1.4.1 After 3 minutes v= 55 km/h = 15,278 mis E =% mv? Ex = % x 4 200kg x (15,278 m/s)?v Ex = 490,176 kiv (2) conrantreseve’ Downloaded from hlayiso.comf=& over

MARIUNG GUIDE: IN| 3 B/OLY(N80)7 14.2 Vezu 15.278 = 0 + a(180) a= 0,085 m/s? S= ut + tat? S= 0 + % x 0,085 m/s? x (3 x 60)? S=1377mv and the height his Sin 2,29°= hy ; 1377 Therefore h = 1377 x Sin 2, 29° h=55, 021 mv Therefore Ep = m.g.h E, = 4 200 x 9,8 x 55,021 E, = 2 264 664,36 J or 2,265 MJv (3) : [16] Copyrightreseved = Downloaded from hlayiso.comcass tum over

BS ° ENGINECRING SCIENCE NS y QUESTION 2: MOMENTS 2.4 The law of moments states that a system of forces is in equilibrium’ if the sum of the left-hand/clockwise moments about the fulcrum is equal to the sum of the right-hand/anticlockwise moments about the same fulcrum.” (2) 2.2 Moments about B XClockwise Moments = ¥Anti-Clockwise Moments (45 x 1) + (30 x 3) +40x5,5=15x05+5x EV E = 69,5 kNw Moments about E ZClockwise Moments = = Anti-Clockwise Moments (40x 0,5)+5xB=30x2+(45x4)+ 15x55” B = 60,5 kN ¥ 50.5kN 45.5kN 24KN -10 kN -29.5KN NN -45.5kN 1 mark each - any four (9) {11] QUESTION 3; FORCES 34 The resultant force of two or more forces is that single force that can replace” the two or more forces and still have the same effect.“ (2) 3.2 4. 150NhasnoVc=0. 150N West 2. 250 N Sin 30° 250 N Cos 30° =125NNorth” = 216,506 N East’ 3. 200 N sin 45° 20 N Cos 45° = 141,421 N South’ = 141,421 N East Vv Please turn over Conwrentreseve’ Downloaded from hlayiso.com

VARIONG C4 3.3 hart} 350 KAA ¢ North 300 NW has Add components DHC = (141,421 N + 216,506 N - 150 N)¥ = 207,927 N East’ 2VC = (300 N + 125 N-141,421 N)Y = 283,579 North” Calculate the resultant R? = (207,927)? + (283,579) R = 351,641 Nv Tana =VC HC Tan a = 283,579 207,927 a=Tan "1, 364 o=53,75°V R= 351.64 N Direction = E 53,76" N/N36,25° Ev Torque =F xr T=F x Cos 30°xr T= 10000 x 0,866 x4 v.. Torque = 8 660 Nm QUESTION 4: FRICTION 44 Copyright reserved e Driving of devices, components and machinery © Bringing linear and rotational movement to halt © Drilling, cutting of materials ° Clamping of workpieces in attachments and jigs e Clamping of workpieces in a vice or chuck (Any appropriate examples 4 x %) STOLE YNGOY (14) (2) [15} (2) Downloaded from hlayiso.commesse um ove

MARA RG GLADELINE 6 T8702 KINSEY: EAGINEERING SCIENCE N3 4.2 4.24 For upwerd movement: F x cosa = 39 200 x sin30° + Fu vw F x cos20° = 39 200 x (0,5) + [0,6(39 200 x Cos30°) — 0,6 x F sin207]v 0,9396 x F = 19 600 + 20 368,9175 — 0,205 x FY 0,9396 x F + 0,205 x F = 39 968.9175 1,144 x F = 39 968,9175v _ 39 968,9175 Lida F = 34 919,55 N or 34,92 kNv (6) 4.2.2 For downward movement: F x cosa + 39 200 x sin30° = Fuv F x cos25° + 39 200 x (0,5) = 0,6(39 200 x Cos30°) + 0,6 x F sin25° 0,906 x F + 19.600 = 20 368,9175 + 0,2536 x FY 0,906 x F : 0,2836 x F = 20 368,9175 ~ 19 600 0,6524 x F = 768,9175¥ “"768,9175 0,6524 F = 1 178,598 N or 1,179 kNY (5) 112] Conrantesere’ Downloaded from hlayiso.cont""""

MARKING GUIDIc YSLOENGO}? QUESTION &: 5.7 The specific heat capacity of a substance is the quantity of energy required to raise the temperature of 1 kg” of a substance by 11°C.” 5.2 Heat gained by water = Heat lost by silver m X she x (tg — t1) = m x she x (ts — f)” 20 kg x 4 187 x (60 - 50) = m x 200 x (120 - 60) 837,400 = mx 12 000W m = 69,78 kg” 5.3 Length = 1,5 m Width = 1m BR34x 10° °C At = (280 ~ 20) At = 260 °Cv Final Area = Ixb +1xbx Bx At Final Area = (1,5 x 1) + (1,5 x 1x 34 x 10%x 260) ¥ Final Area = 1,527 m? v 5.4 5.4.1 Hywet = hf +q x hfg Hwet = 712 + 0,95 x 2 054” Hwet = 712 + 1 951,3 : Hwet = 2 663,3 kd/kgy’ 54.2 — hy=2 766 kskgY 5.4.3 At = (teu — 45) ; At = (200 — 168,3) 168,3 obtain from tables At=31,7°CY © Hey = hg +Cegg X At Hey 2 848,42 kJ/kgy Hes'= Hey - hg Hes = 2 848,42 - 2 766 v Hes = 82,42 kJ/kg’ Copyright reserved Downloaded from hlayiso.com,.,.. turn over (2) {3} (3) (2) (1) (4) [15]

5 MARMING CLE INE 4 BVOEM(NSO)T QUESTION 6: HYDRAULICS 5.4 « Gauge Pressure = pressure shown on a gauge, which excludes outside or air pressure.” « Absolute Pressure = is equal to gauge pressure plus atmospheric pressure.” (2) 6.2 6.2.4 Volume = 27x h Volume = Tr x 0,055? x 0,075¥ Volume = 7,127 x 10% mv 6.2.2 Mass per stroke = 1 m® = 1 000 kg Mass per stroke = 7,127 x 10% x 1.000 v Mass per stroke =0,713 kg¥ 6.2.3 Wad/stroke = F xs Welstroke = 0,713 x 9,81 x 30% Wd/stroke = 209,622 J/stroke’) (3 x2) (6) 6.3 6.3.1 f=F a pb? F = 50 x 10° x (160)"Y (75) F = 227,556 kNV¥, Mass that can be lifted = F = m/g © F = 22 7555556 9,84 F = 23 219,955 kgv (3) 632 _Pr=E Aaa : A A = 3,142 x (0,0375) A= 4,418 x 10° mv Pr=F=50x40°N A 4,418 x 10° Pr= 11,317 MPav (2) [13] Copyright reserved Please turn over Downloaded from hlayiso.com

VARKNG QUIDEL INE QUESTION 7: ELECTRICITY “4 7.4.4 7.1.2 7.1.3 7.44 7.2 7.24 Copyright reserved 4+34+2 Rr = 0,923 OV Rp = 3x0,19 Rb= 0,3 Ov Total Resistance = Rr+ Rb+R4+R5+Re =03+0,923+2+1+01 = 4,323 OV l=V aA 1=6 Volts “ 4,323 |= 1,388 Amps” V=EIxR V=1,388 x1 V = 1,388 VY : Voltage drop of the parallel circuit Vp = 1x Rp Vp =.1,388 x 0,923 Vp.= 1,281 Vv V, = 22 Voltsv Downloaded from hlayiso. FETOUER(NSO)T (4) (2) (1) (2) COMPease turn over

MARONG GUIDELINE “6 4D (IN 2A)T ELEGURICAL TRADE THEORY NS 7.22 Vs = Ip Ve ls ip = Vs x Ig Vp Ip = 22 x2Y 220 Ip = 0,2 AY (2x2) (4) [13] QUESTION 8: CHEMISTRY 8.1 e ron ° Copper * Aluminium e Zinc e Tin e Lead e Gold e Silver ANY SIX ‘mark each (3) 8.2 e Acomposition between metal, oxygen and moisture © Ametal dissolved in an acid e Electrochemical corrosion, which occurs when a electrolytic cell is formed. (Any 2 x 1) (2) 15] TOTAL: 400 Copynahtreseve? Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA EEE eg = ed ee Eee = ————— — —— 0000000000 T590(E)(J24)T AUGUST EXAMINATION | NATIONAL CERTIFICATE. ENGINEERING SCIENCE N3 (15070413) 24 July 2015 (Y-Paper) 13:00-16:00 1 information sheet. Copyright reserved Downloaded from hlayiso.com Please turn over

(16070413) -2- TS90(E\J24)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ENGINEERING SCIENCE N3 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. 2. Answer ALL the questions. Read ALL the questions carefully. Number the answers according to the numbering system used in this question paper. oe Keep subsections of questions together. All the calculations should consist of at 41 The formula used or manipulation thereof 4.2 Substitution of the given data in the formula 43 The answer with the cofrect S! unit The constant values, as they appear on the attached information sheet, must be used wherever possible.” Drawing drawings/ ust be rounded off to THREE decimal places. rite neatly and legibly. Copyright reserved Downloaded from hlayiso.com ?ease tum over

(15070443) -3- T590(E)(J24)T QUESTION 1: MOTION, POWER AND ENERGY 1.4 1.2 1.3 14 1.5 Copyright reserved Sketch and describe displacement of an object (2) Sketch and describe the difference between acceleration and reardation. (2) A racing bike moves from rest and accelerates for 18 secondswith a uniform acceleration of 8 m/s”. Determine the displacement after 18 seconds. (3) A heavy-duty truck with a mass of 18 tons travels at a velocity of 60 | [hy and collides with a bus with a mass of 5 tons travelling in the sam irection at 40 knv/h. After the collision the two objects move togeth direction. Calculate the final common velocity of the TWO obje ‘in m/s? after the collision. F (2x 2) (4) A drilling machine with a pulley diameter of 30.¢m is driven by aflat belt with a thickness of 12 mm. The tension in the tight. ‘side. is 4 320 N and that of the slack side is 1 500 N. The beit velocity is.42 is. Ignore the friction, take the belt thickness into consideration ard calculate the following: ; : 1.5.1 The rotational frequency of the pulley 1.5.2 The power transmitted by the belt in kW ; (2 x 2) (4) [15] Downloaded from hlayiso.com picass tun over

(15070413) -4- TS90(E)(J24)T QUESTION 2: MOMENTS 2.1 Describe the difference between a point load and a uniformly distributed load. 2.2 FIGURE 1 below shows a horizontal beam ABCD of a uniform cross-section, loaded as shown in FIGURE 1. 30 kN 15 kN/im 100 kN Q 2m il 5m 3m il ~ ad nd A B c D FIGURE 1 2.2.1 Calculate by taking moments about be supports. Also test your answer by using Newton’s Third 2.2.2 Draw a shear-force diagram, u suitable scale. Show ALL the main values on the diagram. QUESTION 3: FORCES 3.1 Define triangle of forcé: 3.2 Determine the magnitude and direction of the resultant of the system of forces 1100 N FIGURE 2 (7) (4) [13] (2) (7) Copyrightreseved Downloaded from hlayiso.com Pease tur over

(15070413) -5- TS90(E\J24)T 3.3 Study FIGURE 3 below and answer the questions. 35 kN 12 KN | | f L=?kN R #36 KN FIGURE 3 3.3.1 Calculate the value of L. (2) 3.3.2 Determine graphically or calculat ‘the “magnitude and nature of forces in members PQ and P (4) : [15] QUESTION 4: FRICTION 4.1 (2) 4.2 Weight component perpendicular (F,) to the incline pane (2) Weight component parallel (F.) to the incline plane (2) Smallest force required to move the vehicle up the indine (3) 4.2.4 Smallest force required to move the vehicle down the incline a1} Downloaded from hlayiso.com Copyright reserved Please turn over

(15070413) -6- TS590(E)(J24)T QUESTION 5: HEAT 5.1 Define the specific heat capacity of a substance. (2) 5.2 10 copper components each with a mass of 6 kg are heated up to a temperature of 210°C are then placed in a 9 kg bucket of water at a temperature of 12°C. Calculate the final temperature of the water. (3) 5.3 Determine the change in area of a rectangular brass plate with measurements 200 mm x 600 mm at a temperature of 25°C. The coefficient of. linear expansion is 20 x 10°9/°C if the plate is heated to a temperature of 80.°C. (3) 5.4 A steam boiler developed steam under pressure of 1 450 kPa é is formed with a dryness fraction of 0,9. This wet steam: is now ‘directed to a super heater where it is dried and superheated to a-temperature of 200 °C. The specific heat capacity of superheated steam is 2, ° Klkg°C. Calculate the enthalpy of 3 kg steam of the followi 5.4.1 The wet steam (Hwet) (2) 5.4.2 The dry steam (Hg) Ba (1) 5.4.3 The additional heat required to superheat the steam (Hse). (3) ~ [14] QUESTION 6: HYDRAULICS 6.1 The following da‘ ers to 2 pump which delivers one working stroke per revolution and hich pumps water from a borehole. The area of the pump 4x 10% m’, with a stroke length of 88,42 mm. The pump of 120//min. The vertical depth of the borehole that has to be he following The pressure at which the pump must deliver in kPa The volume of water delivered per stroke of the piston in litres 6.1.3 The power of the pump in kw (3 x 2) (6) Copyright reseved Downloaded from hlayiso.com Pease tum over

(15070413) -7- TSQO(E}(J24)T 6.2 QUESTION 7: ELECTRICITY 7.1 7.2 7.3 Copyright reserved The following data refers to a single-acting hydraulic press. The ratio between the diameter of the plunger piston and the diameter of the ram piston is 1: 4. The hydraulic press displacement of liquid by the plunger piston per stroke is 35 342,917 mm®. Assume the stroke length of the plunger piston to be 60 mm and the force applied to the plunger piston fo be 250 N. Caiculate the following: (3) (2) [14] 6.2.1 The mass in kg that can be raised 6.2.2 The diameter of the plunger piston An electrical circuit consists of three cells connecting: in series, each having an EMF of 4 volts and an internal resistance of;0, 15 ohms, connected in series with a lamp with a resistance of 9 ohms. Calculate the following: 7.1.1 The total EMF of the battery 7.1.2 The total internal resistance of the battery 7.1.3 The total resistance. ofthe circuit 7.14 The current flow in ‘the circuit 7.1.5 The pote tial: di flerence of the battery (5x 1) (5) draws 9 ampere from 240 volts and takes 1,5 minutes to An electrical:ke ( f electricity is 50 cents per kWh. ch will it cost to boil the kettle? (3) How, . surrenit of 15 A is allowed to flow through a solution of silver nitrate. alculate the time required to cover a metal plate with a mass of 150 grams of silver. The electrochemical equivalent for silver is 0.001118 grams per coulomb. (2) Downloaded from hlayiso.com 5.452 tur over

(15070413) -8- TS590(E)(U24)T TA A single transformer has a primary voltage of 220 V and current of 21 A at full load. The secondary current is 7 A. Calculate the following: 741 The turns ratio primary to secondary (2) 7.4.2 The power if the power factor is 0,866 (1) 75 Define ampere. (2) [15] QUESTION 8: CHEMISTRY 8.4 Name TWO methods to combat corrosion. (2) 8.2 Neatly copy and complete the following TABLE in the ANSWER BOOK. COMPOUND eo" ELEMENTS e.g .water Hydrogen-and oxygen Hydrochloric acid : a Table salt Sulphuric acid Limestone (Marble) {4x 1) (4) [6} TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

(15070413) -1- TS90(ENJ24)T ENGINEERING SCIENCE N3 FORMULA SHEET ALL the formulae needed are not necessarily included. Any applicable formula may also be used. W= Fs my Uz tiny .uy=my.v, Lmz V2 =p.Vv W=p D, =(D+) Ww P-— Nnattwet = hy + weg t Pp — os y _ Uitse/Output Ps=latTn.T=Fy 7 = CESCUOUPHE Fog, Inset/Input Ppa Frau _ Fen Ae Fema Aram Apt F, Vram =Vpi xn baw : R Aran - Ara = tanD Fy = Fi 30 Na # Fe # Frsina ...a =0 Fes Esind : Fs = wsin AF y = Ficos0) +... + F,cos0, Fo = weos6 “EF, = FysinO; +... + Fysind, Freosa = F, + Fs ..a=0 R=/SFj + DF FL=T, -T; 2 =1,~Th lano= SEy fi. tension ratio oF y 2 P=F,.v € Q=m.c.dt..tp =tot At mww = QO = inky venr.d.n NV pal 0 t W, AL=Lo.a. At... Lp= Lo AL AE, 2m.g.Ah AA = Ay. B. AL... Ap= Ag AA SER =5.m.aP Qasev-u _ 1 oe O=P.Rut sauttad m=TI.z.t veutat Fp _Np _ Is StF=S1F Vs Ns Ip Downloaded from hlayiso.com Copyright reserved Please turn over

(15070413) -2- TSS0(E)(J24)T 2 CWM = SACWM Paps = Pary + Pyuer P=6xgxh ri Rogp I I map tet ' 1 Rser = Ry + Ry V,-V_=-e(U)-U>) V=IxR Copyright reserved Downloaded from hlayiso.com

(15070413) INFORMATION SHEET TS90(E)(J24)T PHYSICAL CONSTANTS QUANTITY CONSTANTS HOEVEELHEID KONSTANTE Atmospheric pressure 101,3 kPa Atmosferiese dak Density of copper 8 900 kg/m? Digtheid van koyer Density of aluminium 2770 kgim? Digtheid van aluminium Density of gold 49 000 kgim* Digtheid van gout Density of alcohol (ethyl) 790 kgim? Density of mercury 13 600 kg/m" Density of platinum 21 500 kgim* Bigtheld van plaina Density of water 4 000 kgim? Diatheid van waier Density of mineral oil 920 kgim® £ Digtheid van miserale olie Density of air 41,05 kgir 3] Digtheid van lug Etectrochemical equivalent of silver 4,118 mgiC Elektrochemiese ekwivalent van silwer Electrochemical equivalent of copper 0,329 ing/C Elektrochemiese ekwivalent van koper Gravitational acceleration 9,8 m/s- Swaartekragvernelling Heat value of coal °° 30 Mu/kg Warmtewaardeyvan steenkoo! Heat value of anthracite 35 MJ/kg Warmtewaarde van antrasiet Heat value of petrol 45 MJ/kg Warmtewaarde wan petrol Heat value of hydrogen 140 Mu/kg Warmtewaarde ran waterstof Linear coefficient of expansion of copper 47x 10°C Lineére uitsettingskoéffisiént van koper Linear coefficient of ex; i in of aluminium 23 x 10°C Lineére uitsettingskoéffisiént van aluminium Linear coefficient of expansion of steel 12% 10°C Lineére uitsettingskoéffisiént van staal 54x 10°C Lineére uitsettingskoéffisiént van lood 2 100 J/kg.°C | Spesifieke warntekapasiteit van stoom 4187 Jikg.°C | Spesifieke warntekapasiteit van water Specific heat capacity of aluminium 900 J/kg.°C Spesifieke warntekapasiteit van aluminium Specific heat capacity of oil 2 000 J/kg.°C. Spesifieke warrekapasiteit van olie Specific heat capacity of steel 500 d/kg.°C Spesifieke warntekapasiteit van staal Specific heat capacity of copper 390 J/kg.°C Spesifieke warntekapasiteit van koper Copyright reserves Downloaded from hlayiso.com

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higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE NATIONAL CERTIFICATE AUGUST EXAMINATION ENGINEERING SCIENCE N3 24 JULY 2015 This marking guideline consists of 11 pages. Downloaded from hlayiso.com Copyright reserved Please turn over

MARKING GUIDELINE -2- TSGO(E)(J24)T ENGINEERING SCIENCE N3 QUESTION 1: MOTION POWER AND ENERGY 1.1 Bony d c The displacement is the shortest route from the ending to starting point.” (2) 1.2 o acceleration — £ 5 & 3 ; 3 : retardation a i Time in Sec ¥ Acceleration (a) is where the object will increase its velocity and the tetardation/braking is where the object will reduce velocity in that particular face /period.~ (2) 1.3 a= 8 m/s? 0 18 sec S=uxt+hxaxtY S=0x15+ %x8x(18)Y S=1296 mv (3) Copyright reseved Downloaded from hlayiso.comm pease tun over

MARKING GUIDELINE -3- ENGINEERING SCIENCE N3 1.4 M, = 18 tons = 18 000 kg U; = 60 knv/h = 16,667 m/s Mz = 5 tons = 5 000 kg U2 = 40 km/h = 11,111 m/s my, x U4 + me x Us = (m4 + me)V 18 000 x 16,667 + 5 000 x 11,411 = (18 000 + 5 000) V vv V = 300 006 +55 555,556 23 000 v V = 15,459 m/s 1.5 1.5.4 a v= Xen 6 N =60x12 TT X(D +t) N= 60x12 (%) 1(0,3 + 0,012) N = 0,73447 t/s 1.5.2 P=(T-T,)xv P =(4320 ~ 1500) x12 P =33 840 W Copyright reserved or 33,84 kW TSQ0(E)(J24)T (4) (2) (2) [15] Downloaded from hlayiso.com 5.256 irs over

MARKING GUIDELINE -4- T590(E)J24)T ENGINEERING SCIENCE N3 QUESTION 2: MOMENTS 2.1 » Concentrated load is load acting on a point or point load and normally acts vertically downward. e Uniformly distributed load is a load that is spread evenly over the length of a beam or part of the length of a beam. (2x1) (2) 2.2 2.2.1 Moments about A x Anti-Clockwise Moments = = Clockwise Moments (C x 7) = (30 x 2) + (105 x 3,5) + (100 x 10)“Y C = 203,929 kNY Moment about C 2 Anti-Clockwise Moments = = Clockwise Moments (105 x 3,5) + (30 x 5) = (Ax 7) + (100 x 3)v%v A = 31,071 kN ¥ Test Upward = Downward 235 KN = 235 kNY (7) 2.2.2 +100 KN v +100 KN + 31,071 KN v v . OKN 1,071 KN 0 KN v ~ 28,929 kN (4) -103,929 kN [13] Copyright reseved Downloaded from hlayiso.com Pease tum over

MARKING GUIDELINE -5- TS90(E)(U24)T ENGINEERING SCIENCE N3 QUESTION 3: FORCES 3.1 3.2 3.3 Copyright reserved if THREE forces acting on a point are in equilibrium” then they can be represented in magnitude and direction by the sides of a triangle if they are taken in sequence.W (2) The sum of the horizontal components and give the direction > HC =- 700 + 1 100 Cos 25° + 500 x Cos25° v > HC = - 700 + 996,939 + 286,788 ¥ HC = 583,727 kN east’ The sum of the vertical components and give the direction XVC = - 900 - 500 Sin 25 + 1 100 Sin 25 SVC = - 900 — 409,576 + 464,88 v XVC = - 844,696 kN Southy Calculate the resultant R? = (844,696) + (583,727) R= V713511,332 + 340737,211v R= ¥1054248,543 R = 1026,766 kNY Tana=VC HC Tan a = 844,696 583,727 a=Tan “1, 447 a=55° R = 1026,766 KN Direction = E 55° NY (7) 3.3.1 L=12 kN“ (2) 3.3.2 PR = 12/Sin(55) = 14,649 kNv ¥ PQ = 12/Cos55 = 20.921 KN Vv ' 8 Downloaded from hlayiso.com ,,,...4 sin over

MARKING GUIDELINE -6- T590(E)(J24)T ENGINEERING SCIENCE N3 QUESTION 4: FRICTION 41 Brake pads of a vehicle € * Car's tyres on the road e Drilling, cutting of materials e Clutch of a vehicle (4x %) (2) 4.2 4.2.14 Fo=mxg x cosa Fc = 20 000 x cos 18°V Fo = 19 021,130 NY (2) 4.2.2 Parallel force = m x g x sina Fs = 20 000 x sin 18° Fs = 6 180,339 NY (2) 4.2.3 Fup = Fu+Fs Therefore p = Fu mxg y= 2 700 20 000 p = 0,135Y Fu = 0,135 x 19021,130 Fu = 2567,853 NV Fup = Fu + Fs F up = 2 567,853 + 6 180,339 F up = 8 748,192 NV (3) 4.2.4 F down = Fy - Fs F down = 2 567,853 — 6 180,339” F down = -3 612,486 NV (2) 11] Copyrightreseved Downloaded from hlayiso.com Pease tum over

MARKING GUIDELINE -7- TS90(EKI24)T ENGINEERING SCIENCE N3 QUESTION 5: HEAT 5.14 The specific heat capacity of a substance is the quantity of energy required to raise the temperature of 1 kg of substance by 1 °C.” (2) 5.2 Heat gained by water = Heat lost by copper mx she x (T2— 7+) = m x she x (T3 — Ta)” 9kg x 4187 J/kg.°C x (T2 -12) = 60 kg x 402 d/kg.°C x (210 - T2) 37 683 Tz — 454 356 = 5 065 200 — 24120 Tov 37 683 T2 + 24 120 Tz = 5 065 200 + 454 356 61 803 T2 = 5 519 556 Tz = 89,309 °C” (3) 5.3 Length = 600 mm Width = 200 mm a=20x 10°C At = (B0 — 25) At = 55 °Cv New Area=Ixb+2xIxbxaxAt New Area = (0,2 x 0,6) + (2 x 0,2x 0,3 x 20 x 10° x 55V New Area = 0,000132 m°v (3) 6.4 5.4.4 Huet = hf +x hfg Hwet = 838 + 0,9 x 1951” Hwet = 838 + 1 755,9 Hwet = 3 x 2 593.9 kJ/kg Hwet =7 781,7 kd/kgv (2) 5.4.2 Hg = 2 789 x 3 kd/kg = 8 367kJ/kgv (1) 5.4.3 At = (teu — ts) At = (200 — 196,7) 161,4 obtain from tables At = 3,3 °Cv Hey = Hg +Cgg x At Hou = 2 789 + 2,5 x 3,3 Hgy = 2797,25 kJ/kgv Hes = Hey - hg Hes = 2 797,3 - 2 789 Hs = 8,25 x 3 kJ/kg = 24,75 kd/kgv¥ (3) [14] Downloaded from hlayiso.com Copyright reserved Please turn over

MARKING GUIDELINE -8- ENGINEERING SCIENCE N3 QUIESTION 6: HYDRAULICS 6.1 6.1.4 Pr=pxgxh Pr =1 000 x 9,8 x 30¥ P = 196 kPav 6.1.2 Volume = area x height Veaxh V = 26,274 x 10 x 88,42 x 10° V = 2,323 x 104 m3 v But 1 m® = 1 000 litres therefore Volume = 2,323 x 10% x 1 000 Volume = 0,2323 litre’ 6.1.3 P=pxVis or P=Fxs P = 196 x 10° x 120x 10°” P =120x9,8 x 3% 60 P = 392 WY P =392 WY 6.2 6.2.1 f= F d? BD? F= fxD? a F = 250 x(4)°v (1° F = 4000 NY m=W 9,8 m = 4.000 9.8 m = 408,163 kgv Copyright reserved Downloaded from hlayiso.com Pease tum over T590(E)(J24)T (3x 2) (8) (3)

MARKING GUIDELINE 6.2.2 -g- ENGINEERING SCIENCE N3 Volume = area x length Area = Volume = 35 342,917 Length 60 A = 589,048 mm?v But the Area = 1d? 4 D= /areax4 T D= /§589,048 x 4 7 D = 27,384 mmv QUESTION 7: ELECTRICITY 7.1 7.1.1 7.1.2 7.1.3 Copyright reserved EMF =3x4V =12Vv¥ Rr =3 x 0,150 = 0,45 OY Ry; =Rtr =9+0,45 = 9,45 OY 9,45 |= 1,2698 AY = F< V=Rx! 9x 1,2698 1 1,429 volts¥ Vv Downloaded from hlayiso.com T5O0(E)(J24)T (2) [11] (5x 1) Please turn over (5)

MARKING GUIDELINE 7.2 7.3 74 7.5 Copyright reserved Wd=PxVxt = 240 x9 x90 Wd = 194 400 Jv 3,6 MJ = 50° cost = 194 400 x 50V” 3,6 x 10 Cost = 2,7 cents “ me=txfxt t=m Ixf = 1509 1,118x10° x 15” -10- ENGINEERING SCIENCE N3 t = 8944.544 s/149,08 min/2,18 hours” 7.4.1 No = ds Ns Ip =f 21v No #1 Ns 3 No iNs = 1:3 7.4.2 Power = VI cos@ = 220 x 21 x 0,866 = 4 000,92 W Power = 4 kW Ampere is the voltage multiplied with the current~ TS90(E)(J24)T (3) (2) (2) (1) (2) [15] Downloaded from hlayiso.com Pease turn over

MARKING GUIDELINE T590(E)(J24)T ENGINEERING SCIENCE N3 QUESTION 8: CHEMISTRY 8.1 8.2 Copyright reserved e By painting the metal e By applying oil/grease e By electroplating with zinc/tin (Any 2 x 1) COMPOUND TWO COMPOUND ELEMENTS Hydrochloric acid Hydrogen, chlorine and oxygen’) Table salt Sodium and chloride” Sulphuric acid Hydrogen, sulphur and oxygen” Limestone (Marble) Calcium, carbon and oxygen” TOTAL: Downloaded from hlayiso.com (2) (4) [6] 100

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