CONFIDENTIALUNIVERSITI TUN HUSSEIN ONN MALAYSIA ANSWER SCHEME FINAL EXAMINATION SEMESTER I SESSION 2014/2015 COURSE NAME : FLUID MECHANICS COURSE CODE : BFC10403/BFC1043 PROGRAMME : 1 BFF EXAMINATION DATE : JANUARY 2015 DURATION INSTRUCTION : : 3 HOURS ANSWER FOUR (4) FROM FIVE (5) QUESTIONS CONFIDENTIAL THIS QUESTION PAPER CONSISTS OF EIGHT (8) PAGES Q1 (a) Determine the specific weight, density, specific volume and specific gravity of certain liquid with volume 6.5 m3 and weight 55kN. BFC 10403/BFC 1043 (12 marks) 3 m W 55 103 8461.5 N/m3 specific weight, = g V V 6.5 γ 8461.5 density, = g 9.81 862.5 kg/m3 1 3 1 specific volume ,v = ρ 862.5 1.1594 10-3 m3/kg ρ cecair 862.5 0.8625 specific gravity, sg = ρ air 1000 3 3 (b) Explain briefly and sketch a capillarity effect and surface tension. 2 The phenomenon of capillary effect can be explained microscopically by considering cohesive forces (the forces between like molecules, such as water and water) and adhesive forces (the forces between unlike molecules, such as water and glass). 2 2 BFC 10403/BFC 1043 2 The pulling force that causes this tension acts parallel to the surface and is due to the attractive forces . Between the molecules of the liquid. The magnitude of this force per unit length is called surface tension s and is usually expressed in the unit N/m 2 (8 marks) (c) Calculate the minimum diameter of a glass tube if a capillary rise is not more than 0.25 mm. Given, the surface tension = 0.075 N/m and specific weight of water , = 9810 kg/m3 . ( 5 marks) Given, h = 0.25 mm = 2.5 × 10– 4 m surface tension , = 0.075 N/m specific weight of water , = 9810 kg/m3 . Capillary rise, h = 2 4 cos d As we know , = 0, 2 2.5 × 10 –4 4 0.075 = 9810 d 1 d = 0.122 m = 122 mm 3 BFC 10403/BFC 1043 Q2 (a) Determine the absolute pressure in unit kPa if barometer reads 60 kPa. Given, barometer height at sea level is 740 mmHg and sgmerkuri = 13.6 . ( 8 marks) Diberi:Gage pressure, Pgage = 60 kPa Barometer height = 740 mmHg So, Patm = gh 1 740 2 1000 P (13.6 1000) (9.81) Patm = 98.728 kPa And, 1 2 Pabsolute = Pgage + Patm Pabsolute = 60 + 98.728 2 Pabsolute = 158.728 kPa (b) Calculate the pressure difference (PB-PA) in double–fluid manometer as shown in Figure Q2(b).. (13 marks) 4 BFC 10403/BFC 1043 Properties The specific gravities are given to be 13.5 for mercury, 1.26 for glycerin, and 0.88 for oil. density of water =1000 kg/m3 2 4 3 2 2 (c) Explain why the pressure in fluid increases with depth? (4 marks) Water is NOT incompressible. As you go deeper, its density increases with pressure. Before you go down, there is very little room between the molecules. But you go down 30 m, there is now even less room. This means that the number of collisions will go up dramatically. So the pressure increases. 4 5 1 1 2 1 BFC 10403/BFC 1043 Q3 (a) (i) (ii) State TWO (2) differences between laminar and turbulent flow, and function of the Reynolds number in the pipe. If the Reynolds number of mercury is 1.789 x 103, what type of flow can you classify and why? (6 marks) Laminar flow Re < 2000 'low' velocity Dye does not mix with water Fluid particles move in straight lines Simple mathematical analysis possible Rare in practice in water systems 2 Turbulent flow Re > 4000 'high' velocity Dye mixes rapidly and completely Particle paths completely irregular Average motion is in the direction of the flow Cannot be seen by the naked eye Changes/fluctuations are very difficult to detect. Must use laser. 2 (ii) Turbulent : Re > 4000 6 2 BFC 10403/BFC 1043 (b) Fluid A flows through a galvanised iron pipe with length and diameter are 45 m and 150 mm respectively for discharge 5.630 L/s. Calculate the head loss due to fluid friction of pipe. (Given = 869 kg/m3, = 8.14 x 10-2 Pa.s, 1 m3 = 1000 L). (8marks) L = 45.0 m , D = 150 mm = 0.15 m Q = 5.630 L/s Find hf ? V Re Q 5.63 x10 3 0.32m / s A 0.15 2 4 2 VD 869 x0.32 x0.15 512.4 2,000 ( lamina) 2 8.14 x10 2 f 64 64 0.125 NR 512.4 2 L v hf f D 2g 2 L v2 45 (0.32) 2 (0.125) 0.1957 m 2 D 2g 0.15 2(9.81) hf f 7 BFC 10403/BFC 1043 (c) Figure Q3(c) shows a pressurized tank of water has a 10-cmdiameter orifice at the bottom, where water discharges to the atmosphere. The water level is 3m above the outlet. The tank air pressure above the water level is 300 kPa while the atmospheric pressure is 100 kPa. Neglecting frictional effects, determine the initial discharge rate of water from the tank. (11 marks) 8 BFC 10403/BFC 1043 3 4 2 2 Q4 (a) Briefly explain the characteristics of discharge Q and head loss hf of flow in pipes installed as follows; (i) (ii) Series Parallel (7 marks) Series Pipe 1 Pipe 2 1.5 QT = Qpipe 1 = Qpipe 2 2 hL = hL pipe 1 + hL pipe 2 Parallel 1.5 QT = Qpipe 1 + Qpipe 2 9 hL = hL pipe 1 = hL hf pipe 2 BFC 10403/BFC 1043 2 Pipe 1 Pipe 2 (b ) List ONE (1) major head loss and THREE (3) minor head losses incurred in a water distribution system (4 marks) 1 Major head loss – hf = 3 Minor head loss – hm = (c) , A flow of water has been discharge through a horizontal pipeline to the atmosphere. The pipeline is connected in series and consisted of two pipes which are 10 cm diameter and 25 m long and 12 cm diameter and 35 m long. The friction factor is 0.002 for both pipes. The water level in the tank is 10 m above the centre-line of the pipe at the entrance. Considering all the head losses, calculate the discharge when the 10 cm diameter pipe is connected to the tank ΦA= (13 marks) Q pipe 1 = Q pipe 2 0.5 V1A1 = V2A2 0.5 V1 D12 = V2 D22 V2 = V1 A12 / A22 = 0.694 V1 0.5 10 BFC 10403/BFC 1043 Loss at entrance, hm 0.5 V1 2 2g 0. (V V 2 ) 2 Loss at expansion, hm 5 1 2g 1 2 hm V1 (1 0.694) 2 0.0934V1 2 2g 2g Head loss in 10 cm pipe due to friction, V fLV1 2 (0.002)(252 )V12 0.5 1 2g 2 gD 2 g (0.1) 2 h f1 Head loss in 12 cm pipe due to friction, 1 h f2 2 V fLV 2 2 (0.002)(35)(0.694) 2 V12 1.69 1 2g 2 gD 2 g (0.12) 2 (0.5 0.0934 0.5 1.69)V1 2 V 2.783 1 Total head loss = h L 2g 2g p1 v12 p v2 z1 2 2 z 22 h L 2g 2g 10 0.694 V12 V2 V2 2.783 1 3.477 1 2g 21g 2g V1 1 7.51m / s (0.12 ) 0.059m 3 / s 59 L / s 1 4 Q VA 7.51x Q5 (a) Differentiate between model and prototype. (4 marks) 11 BFC 10403/BFC 1043 Model : Prototype: (b) Similar with object/structure required in certain scale ratio. tested in laboratory and similar in real phenomenon. 2 not necessary its smaller than prototype object/actual structure tested in actual phenomenon, example: structure in open 2 channel, ship etc Clearly explain the differences between geometry and kinematics similitude. (7 marks) Geometry similitude Kinematics similitude The prototype and model have identical shapes but differ only in size. 1 1 In addition to Geometric Similarity, ratio of velocities at all corresponding points in flow are the same 12 BFC 10403/BFC 1043 Ratio of corresponding length in prototype and model show as, Lenght = Lp LM LR Involve length and time. Time = Tp TM TR 2.52. 5 Area = Ap AM Vp LP 2 LM 2 LR 2 LP LP Vp T L L P M R Velocity = TP V M LM TR TP TM 3 L 3 P 3 LR Volume = V M LM (c) Derived an equation of non-dimension group by using Buckingham Theorem to describe the resistance force (F). The resistance force (F) for a ship influenced by the function length L, velocity V, acceleration gravity g, density flow ρ and dynamic viscosity μ. (Repeating variables : L, V and ρ ) (14 marks) 1 13 BFC 10403/BFC 1043 Given : F = fn (L, V, g, ρ, µ) and repeating variables : L, V dan ρ F = fn (L, V, g, ρ, µ) which f – fungsi and n = 6 ML L L M M f ( L, , 2 , 3 , ) 1 T T L LT T2 m = 3 (M, L and T) & n-m = 6-3 = 3 3 Pi that will find is ( 1, 2, 3) = 0 Repeating variables L, V dan ρ 1 = La1 V2b1 c1 F 2 = La2 Vb2 c2 g 3 = La3 Vb3 c3 µ For 1 1 = La1 Vb1 c1 F b1 c1 M0 L0 T0 = L a1 LT 1 ML3 MLT 2 For M : 0 = c1 + 1 2 For L : 0 = a1 + b1 -3c1 +1 For T : 0 = - b1 - 2 a1 = -2, b1 = -2, c1 = -1 1 = F V 2 L2 For 2 2 = La2 Vb2 c2 g 2c1 b1 M0 L0 T0 = L a1 LT 1 ML3 LT 2 For M : 0 = c2 For L : 0 = a1 + b1 - 3c1 + 1 For T : 0 = - b1 - 2 a1 = 1, b1 = -2, c1 = 0 gL 2 = 2 V For 3 2 3 = La3 Vb3 c3 µ b1 c1 M0 L0 T0 = L a1 LT 1 ML3 ML1T 1 For M : 0 = c1 + 1 2 For L : 0 = a1 + b1 -3c1 - 1 For T : 0 = - b1 - 1 a1 = -1, b1 = -1, c1 = -1 3 = VL 2 F gL , ] 2 2 = fn [ V L V 2 VL 14 BFC 10403/BFC 1043 15