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Unformatted text preview: University of California Davis Department of Mechanical and Aerospace Engineering ENG 103 Fluid Mechanics Fall 2011 Professor Kennedy Homework 2 Solutions Chapter 3 Integral Relations for a Control Volume P3.13 The cylindrical container in Fig. P3.13 is 20 cm in diameter and has a conical contraction at the bottom with an exit hole 3 cm in diameter. The tank contains fresh water at standard sea-level conditions. If the water surface is falling at the 183 h(t) D nearly steady rate dh/dt 0.072 m/s, estimate the average velocity V from the bottom exit. Fig. P3.13 V? Solution: We could simply note that dh/dt is the same as the water velocity at the surface and use Q 1 = Q 2 , or, more instructive, approach it as a control volume problem. Let the control volume encompass the entire container. Then the mass relation is dm | system dt or : 4 D2 0 d ( d) dt CV dh dt 4 Introduce the data : mout 2 Dexit V V ( d ( dt 0 Cancel cone 4 4 :V 20 cm 2 m ) [ ( 0.072 )] 3 cm s D 2 h) | ( 4 D 2 dh )( ) Dexit dt 3.2 m s P3.14 The open tank in the figure contains water at 20 C. For incompressible flow, (a) derive an analytic expression for dh/dt in terms of (Q 1 , Q 2 , Q 3 ). (b) If h is constant, determine V2 for the given data if V 1 3 m/s and Q 3 0.01 m3/s. Solution: For a control volume enclosing the tank, d dt d (Q2 Q1 Q3 ) d 2 dh 4 dt 2 Dexit V , (Q2 Q1 Q3), Ans. s solve V jet 6.06 m s 4 Ans. P3.33 In some wind tunnels the test section is perforated to suck out fluid and provide a thin viscous boundary layer. The test section wall in Fig. P3.33 contains 1200 holes of 5mm diameter each per square meter of wall area. The suction velocity through each hole is V r 8 m/s, and the test-section entrance velocity is V 1 35 m/s. Assuming incompressible steady flow of air at 20°C, compute (a) V o , (b) V 2 , and (c) V f , in m/s. Fig. P3.33 Solution: The test section wall area is ( )(0.8 m)(4 m) 10.053 m2, hence the total number of holes is (1200)(10.053) 12064 holes. The total suction flow leaving is Qsuction NQ hole (12064)( /4)(0.005 m)2 (8 m/s) 1.895 m 3 /s (a) Find Vo : Q o Q1 or Vo solve for Vo (b) Q 2 Q1 Qsuction or: V2 (c) Find Vf : Q f 4 m 3.58 s (35) 4 31.2 Q2 solve for Vf (2.5)2 (35) Ans. (a) (0.8)2 1.895 V2 m s or Vf 4 (0.8)2, 4 (0.8)2 , Ans. (b) (2.2)2 4 m 4.13 s (31.2) Ans. (c) 4 (0.8)2 , dt dt As the disk spacing drops, h(t) V o t, the outlet velocity is V ho V o r/(2h). Ans. P3.39 A wedge splits a sheet of 20 C water, as shown in Fig. P3.39. Both wedge and sheet are very long into the paper. If the force required to hold the wedge stationary is F = 126 N per meter of depth into the paper, what is the angle of the wedge? 6 m/s 6 m/s F Fig. P3.39 4 cm Solution: For water take m/b Vh 6 m/s = 998 kg/m3. First compute the mass flow per unit depth: (998 kg / m3 )(6 m / s )(0.04 m) Chapter 3 239.5 kg / s m Integral Relations for a Control Volume 201 The mass flow (and velocity) are the same entering and leaving. Let the control volume surround the wedge. Then the x-momentum integral relation becomes Fx or : F m(uout 124 N / m Solve cos 2 uin ) m (V cos 2 V) mV (cos (239.5 kg / s m)(6 m / s)(cos 0.9137 , 2 24o , P3.40 The water jet in Fig. P3.40 strikes normal to a fixed plate. Neglect gravity and friction, and compute the force F in newtons required to hold the plate fixed. Solution: For a CV enclosing the plate and the impinging jet, we obtain: 48o 2 1) Ans. 2 1) in the x-direction is: Fx Chapter 3 Integral Relations for a Control Volume Fbolts p1A1 mu2 mu1 m( V2 cos 40 295 V1 ) Solution: (a) With(2.3 turbine, “1” (0.1)2 15.25(21.7 cos 40 1.95) solve for Fbolts the 101350) is upstream: 4 2 p1 V1 p2 V 2 2 z1 z2 h f h t, g 2g vane ong 2g P3.61 A 20°C water jet strikes a 2100 N Ans. a tank with frictionless wheels, as shown. or: 0 without The jet turns and falls into the tank 0 150 0 0 25 17 h t 30°, estimate the horizontal spilling. If 3 fSolve for h t 108 ft. Convert Q 15000 gal/min 33.4 ft /s. Then the turbine power is orce F needed to hold the tank stationary. ft lbf 410 hp Ans. (a) Solution: PThe Qh turb (62.4)(33.4)(108) 225,000 CV surrounds the tank and s wheels and cuts through the jet, as shown. We For pump operation, point “2” is upstream: Fig. P3.61 (b) should assume that the splashing into the tank does not increase the x-momentum 2 of the water in the tank. Then we2can write1 V1 p2 V 2 p z2 z1 h f h p , the CV horizontal force relation: g 2g g 2g d Fx F u d 25 m in u in 0 mVjet independent of dtor: 0 0 tank 0 0 150 17 h p 2 2 slug 2 Solve for h p 142 ft ft 106 lbf Ans. Thus F A jV 2 1.94 3 ft 50 j 4 12 s ft Qhp (62.4)(33.4)(142) 296000 ft·lbf/s 540 hp. Ans. The pump power is Ppump (b) Or do the question in SI units to the standard P3.62 Water at 20°C exits sea-level atmosphere through the split nP3.175 inWFig. at 20°C is delivered from one reservoir to another through a long 8-cmozzle ater P3.62. Duct areas are 2 A 1 0.02pipe.aTheA 2 Areservoir has 2. surface elevation z 2 80 m. The friction loss in diameter m nd lower 3 0.008 m a If pt1 pipe is correlated byathe the flow rate is 17.5(V2/2g), where V is the average velocity he 135 kPa (absolute) nd formula h loss 75 m steady flow the force on Qn the pipe.2If the 3/h, computerate through the pipe is 500 gallons per minute, estimate the i 2 Q3 the flange bolts at of the higher reservoir. surface elevation section 1. Solution: We may apply Bernoulli here. Convert 500 gal/minig. 0.0315 m3/s. F = P3.62 hf 17.5V 2 2g z1 z2 0.0315 m3 / s 17.5 2(9.81 m/s 2 ) ( / 4)(0.08m) 2 z1 115 m 2 z1 80 m Ans. Q 6. A 10 mm diameter jet of water is deflected by a homogeneous rectangular block (15 mm by 200 mm by 100 mm) that weighs 6 N as shown in this figure. Determine the minimum volume flow rate needed to tip the block over. ...
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This note was uploaded on 12/18/2011 for the course ENG 103 taught by Professor Chattot during the Fall '08 term at UC Davis.

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