For the situation shown, determine the minimum possible ...
TEST 2/ EXAM, FLUID MECHANICS FOR W, VVR12023 OCTOBER 2010, 08:00-11:00 (Test 2), 08:00-13:00 (Exam)Open book: Text book, lecture notes, collection of equations, dictionary, pocket calculator.Remember: Answers should be short and concise, structured and in punctual form. Just one solved question on each paper. Calculations should be clear with assumptions and units. Write your name on each paper.General: Use (if no other information given) for water: ??20oC) =1.0103 kg/m3, ? = 10-3 Pas, g = 9.81 m/s2, patm = 100 kPa.Exam results: Examination results will be displayed 10 November, 2010.Grading: Given on basis of 2 written tests. More information in course program.TEST 2 CONSISTS OF 4 PROBLEMS (TOTALLY 36 POINTS)EXAM CONSISTS OF 8 PROBLEMS (TOTALLY 60 POINTS)TEST 1 (24 POINTS)---------------------------------------------------------------------------------------------------------------------PROBLEM 1 (4 points)A lubricant between a piston and cylinder according to below figure has a kinematic viscosity of 2.8?10-5 m2/s and a relative density of 0.92. If the piston has a mean velocity of 6 m/s what approximate power loss will develop due to friction. The piston has a diameter of 150 mm and length 300 mm. The cylinder has a diameter of 150.2 mm.150 mm150.2 mm300 mmLubricantPROBLEM 2 (4 points)A regulated lake with an area of 5 km2 has a water level of +11.07 at 07:00 am. The only inflow to the lake is a constant 6 m3/s and the only outflow is through a power plant. How will the water level vary during a day (24 hours) if 12 m3/s discharge through the power plant between 07:00 am and 19:00 pm and 2 m3/s discharge during the rest of the time?PROBLEM 3 (6 points)The fire brigade has developed a new high pressure nozzle to be used against aggressive demonstrators at the next G7 meeting in Lund. The water speed is 20 m/s just at the 0.04 m diameter nozzle outlet. How large force does each still standing demonstrator need to develop in order not to be removed by the horizontal water jet? State all necessary assumptions clearly.PROBLEM 4 (10 points)The rigid gate OAB in below figure is hinged at O and rests against a rigid support at B. What minimum horizontal force P is required to hold the gate closed if its width is 5 m? Neglect the weight of the gate and friction in the hinge. To the right and under the gate is atmosphere.WaterP4 m3 m2 mHingeABOAtmosphereWRITTEN TEST 2 (36 POINTS)---------------------------------------------------------------------------------------------------------------------PROBLEM 5 (10 points)Determine the flows in the pipes between the four reservoirs in below figure. Pipe data are shown in the table and all pipes have a friction coefficient of f = 0.02. (El. = elevation in m above reference level).PipeLength (m)Diameter (mm)(1)200500(2)600300(3)1500300(4)1500400(1)El. 30 m(2)(3)(4)El. 250 mEl. 300 mEl. 200 mPROBLEM 6 (10 points)A urine separating toilet has been installed. When the toilet tank is full the contents can be spread to the ground. The tank is constructed through three steel plates (1.2 m x 1.2 m), out of these, one has been divided along the diagonal so that the sides constitute two triangles (without lid according to below figure, the tank is 1,2 m long in to the paper). The tank top is covered by a plastic with holes. In the tank bottom five holes are connected to five hoses of equal properties (2 m long, 12 mm diameter, with friction coefficient f = 0.05). Assume local loss at hose entrance k=0.8 and the hose bend k=0.3. The tank bottom is 0.5 m above the ground. Determine how long time it will take to empty the tank if it is completely full from start. Assume that the contents have the same properties as water. Disregard acceleration effects. In below figure the tank seen from the side, note that just one of the five hoses is seen in the figure.1,7 m1,2 mHosePROBLEM 7 (8 points)A channel has a slope of 1 ‰. The section is trapezoidal with the bottom width equal to 1 m and sides sloping 1:2. The channel material is cement mortar. What depth must the channel have to allow a uniform discharge of 2 m3/s?PROBLEM 8 (8 points)A submarine has the shape of an 8:1 ellipsoid. The water temperature is 20 C. The submarine has a frontal area of Afront = 4.64 m2 and a drag coefficient Cd = 0.15.a) Find the power required to maintain a velocity of 6.1 m/s fully submerged in the sea. b) What is a likely method that has been used to find the value of Cd?c) Now, assume that the value of the drag coefficient is not given. Instead you make an estimate based on the approximation that the resistance is due to friction only. What is then the total drag force? Assume that the shape of the submarine is a perfect cylinder with length L = 50 m.Solutions1. ρ = r.d. ? 1000 = 920 kg/m3; μ = υ?ρ = 2.8?10-5?920 = 0.0258 Pa?s; τ = μ(V/h) = 0.028?(6/0.0001) = 1548 Pa. The friction force F on the piston is F = τ?A= 1548?π?0.15?0.3 = 219 N. The power loss P due to friction is P = F?V = 219?6 = 1.313 kW.2. The continuity equation: dS/dt = Qin-Qout. During 07-19 continuity gives: dS/dt = (6-12) = -6 m3/s. Storage change dS = -6?12?3600 = -259 000 m3. This corresponds to change in level -259 000/(5?106) = -0.052 m. At 19 the lake level is then +11.07-0.052 = +11.018 m. During 19:00 to 07:00 the next day dS/dt = (6-2) = 4 m3/s. This gives dS = 172 800 m3. The corresponding lake water level will be +11.018 + 0.035 = +11.053 m. Consequently, assuming no edge effects and no other losses or inflows, the lake water level will vary in a linear fashion between +11.07 m at 7 am to +11.018 m at 19 pm. Next morning at 7 am the water level will be +11.053 m and then decrease to +11.001 at the following 19 pm.3. Assume that the horizontal water jet changes direction so that after hitting the demonstrator it will have a perpendicular direction from the original one. The x-direction is along the horizontal water jet and the y-direction is perpendicular to this. Use the momentum equation for a control volume between a section A just at the outlet from the nozzle and section B just at the demonstrator body. The force Px on the control volume isPx = ρQVBx - ρQVAx = 0-103?20?0.022?π?20 = -503 N. Consequently, the demonstrator is affected by 503 N from the water and needs to develop a similar force in the other direction. The force corresponds to a weight of about 50 kg!4. Fh = pressure force on gate side OA; Fv = pressure force on gate side AB; lh = lever arm for Fh about hinge; lv = lever arm for force Fv about hinge; yc,OA, yc,AB = are vertical distances from water surface to center of gate side OA and AB, respectively; yp,OA = vertical distance from water surfaceto center of pressure on gate side OA; Ic,OA = second moment of area for gate side OA about its center; AOA, AAB = surface areas for gate side OA and AB, respectively.yc,OA = 4+1.5 = 5.5 myc,AB = 4+3 = 7 mAOA = 3?5 = 15 m2AAB = 2?5 = 10 m2Ic,OA = 5?3/12 = 11.25 m4yp,OA = yc,OA + Ic,OA/( yc,OA? AOA) = 5.636 mFh = ρH20? yc,OA ? AOA = 809.3 kNFv = ρH20? yc,AB ? AAB = 686.7 kNlh = (yc,OA – 4 = 1.636 mlv = 1 mThe moment about the hinge then becomes:Pmin?3 = Fh ? lh + Fv ? lv => Pmin = 670 kN.5. Assign reservoir El. 30 by A; El. 250 by B; El. 300 by C; El. 200 by D; and the joint between all pipes by J. The general friction equation gives:hf1 = 0.02?200/0.55?16?Q21/(2g?π2) = 10.6?Q21; hf2 = 408.1?Q22; hf3 = 1020.4?Q23; hf4 = 242.1?Q24;Assume flow from reservoir B, C, and D to reservoir A. Denote the energy level in J by Hj.Energy eqn B => J: 250 = Hj+hf2 <=> 250 = Hj+408.1?Q22 (1)Energy eqn C => J: 300 = Hj+hf3 <=> 300 = Hj+1020.4?Q23 (2)Energy eqn D => J: 200 = Hj+hf4 <=> 200 = Hj+242.1?Q24 (3)Energy eqn J => A: Hj = 30+hf1 <=> 250 = Hj = 30+10.6?Q21 (4)Continuity Q1=Q2+Q3+Q4 (5)Solve the above equation system by guessing Hj (should be between 30 and 200 m if the assumed flow direction is true), calculate Q1, Q2, Q3, and Q4 by eqn (1) to (4), check with eqn (5). This gives Hj = 67.7 and Q1= 1890 l/s, Q2 = 670 l/s, Q3 = 480 l/s and Q4 = 740 l/s.6. The height h of the tank is calculated from 1.72/22 + h2 = 1.22 => h = 0.85 m. The water surface area A varies depending on the water level A = a?b where a is constant 1.2 m and b varies according to b/z = 1.7/0.85 = 2; b = 2?z and A = 1.2?2?z = 2.4?z. Assume point 1 on top of water surface and point 2 in pipe center at discharge point. The vertical distance between 1 and tank bottom is z = 0.494 m (0.5-0.012/2 ≈ 0.5). The time it takes to change water level from z2 to z1 from a tank with varying water level is calculated by ∫z2z1 A?dz/(0-Qout) (assuming no inflow). z1=0 and z2=0.85 m. Qout is a function of the water level in the tank. Assume quasistationarity for flow. Energy equation between 1 and 2 (p1=p2=0, z1=z; V1=0); 0+0+z=V22/(2?g) – 0494 + (0.3+0.8+0.05?2/0.012) ? V22/(2?g) => V2 = 1.37?√(z+0.494)Qout=5?V2?Ap where Ap = area of pipe section. This gives Qout = 0.000775?√(z+0.494) thent = ∫0.850 (2.4?z?dz)/(-0.000775?√(z+0.494)) => t = -3097.5 ? ∫0.850 (z?dz)/(√(z+0.494)) This gives t = -3097.5 ? (2/3?(z+0.494)1.5 – 0.988?(z+0.494)0.5)00.85 => t=1103 s = 18 min.7. Manning gives: Q = So1/2?A?R2/3/n; So = 1/1000; Table 8.1 in text book gives n = 0.013 for cement mortar. A = 1.0?y+2?1/2?2?y?y where y is the required depth. P = 1.0 + 2?y?√(1+22); R = (y+2y2)/(1+2?y?√5). This gives the flow:Q = 0.0011/2?(1.0?y+2?y2) ? ((1.0?y+2?y2)/((1+2?y?√5)))2/3/0.013. Trial and error for Q=2 gives y≈0.6 m.8. ................
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