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the bellows and the outlet flow, we obtain Fig. P3.31 2 out d ( ) m 0, where bhL bL tan dt \u3c1\u3c5 \u3c5 \u3b8+ = = =\ufffd 2 o d since L is constant, solve for m ( bL tan ) . dt Ans\u3c1 \u3b8= \u2212 = \u2212\ufffd \u3b8\u3c1 \u3b82 2 dbL sec dt 3.32 Water at 20°C flows through the piping junction in the figure, entering section 1 at 20 gal/min. The average velocity at section 2 is 2.5 m/s. A portion of the flow is diverted through the showerhead, which contains 100 holes of 1-mm diameter. Assuming uniform shower flow, estimate the exit velocity from the showerhead jets. Solution: A control volume around sections (1, 2, 3) yields 3 1 2 3Q Q Q 20 gal/min 0.001262 m /s.= + = = Meanwhile, with V2 = 2.5 m/s known, we can calculate Q2 and then Q3: 3 2 2 2 2 m(2.5 m) (0.02 m) 0.000785 , 4 s Q V A \u3c0= = = 164 Solutions Manual \u2022 Fluid Mechanics, Fifth Edition 3 3 1 2 m0.001262 0.000785 0.000476 s hence Q Q Q= \u2212 = \u2212 = 3 2 3 m/100 0.00000476 (0.001) , s 4 . jet jet Each hole carries Q V solve V Ans \u3c0 = = = m6.06 s 3.33 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 5-mm diameter each per square meter of wall area. The suction velocity through each hole is Vr = 8 m/s, and the test-section entrance velocity is V1 = 35 m/s. Assuming incompressible steady flow of air at 20°C, compute (a) Vo, (b) V2, and (c) Vf, in m/s. Fig. P3.33 Solution: The test section wall area is (\u3c0)(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 2 3 suction holeQ NQ (12064)( /4)(0.005 m) (8 m/s) 1.895 m /s\u3c0= = \u2248 2 2 o o 1 o o (a) Find V : Q Q or V (2.5) (35) (0.8) , 4 4 solve for V . (a)Ans \u3c0 \u3c0 = = \u2248 m3.58 s 2 2 2 1 suction 2 2 (b) Q Q Q (35) (0.8) 1.895 V (0.8) , 4 4 or: V . (b)Ans \u3c0 \u3c0 = \u2212 = \u2212 = \u2248 m31.2 s 2 2 f f 2 f f (c) Find V : Q Q or V (2.2) (31.2) (0.8) , 4 4 solve for V . (c)Ans \u3c0 \u3c0 = = \u2248 m4.13 s Chapter 3 \u2022 Integral Relations for a Control Volume 165 3.34 A rocket motor is operating steadily, as shown in Fig. P3.34. The products of combustion flowing out the exhaust nozzle approximate a perfect gas with a molecular weight of 28. For the given conditions calculate V2 in ft/s. Solution: Exit gas: Molecular weight = 28, thus Rgas = 49700/28 = 1775 ft2/(s2\u22c5°R). Then, Fig. P3.34 2 exit gas p 15(144) psf 0.000780 slug/ft RT (1775)(1100 460)\u3c1 = = \u2248+ For mass conservation, the exit mass flow must equal fuel + oxygen entering = 0.6 slug/s: \u3c0\u3c1 \ufffd \ufffd= = = \u2248\ufffd \ufffd \ufffd \ufffd \ufffd 2 exit e e e e e slug 5.5 m 0.6 A V (0.00078) V , solve for V . s 4 12 Ansft4660 s 3.35 In contrast to the liquid rocket in Fig. P3.34, the solid-propellant rocket in Fig. P3.35 is self-contained and has no entrance ducts. Using a control-volume analysis for the conditions shown in Fig. P3.35, compute the rate of mass loss of the propellant, assuming that the exit gas has a molecular weight of 28. Fig. P3.35 Solution: With M = 28, R = 8313/28 =297 m2/(s2\u22c5K), hence the exit gas density is 3 exit p 90,000 Pa 0.404 kg/m RT (297)(750 K)\u3c1 = = = For a control volume enclosing the rocket engine and the outlet flow, we obtain CV out d (m ) m 0, dt + =\ufffd 2 propellant exit e e e d or: (m ) m A V (0.404)( /4)(0.18) (1150) . dt Ans\u3c1 \u3c0= \u2212 = \u2212 = \u2212 \u2248\ufffd \u2212 kg11.8 s 166 Solutions Manual \u2022 Fluid Mechanics, Fifth Edition 3.36 The jet pump in Fig. P3.36 injects water at U1 = 40 m/s through a 3-in pipe and entrains a secondary flow of water U2 = 3 m/s in the annular region around the small pipe. The two flows become fully mixed down- stream, where U3 is approximately constant. For steady incompressible flow, compute U3 in m/s. Solution: First modify the units: D1 = 3 in = 0.0762 m, D2 = 10 in = 0.254 m. For incompressible flow, the volume flows at inlet and exit must match: 2 2 2 2 1 2 3 3Q Q Q , or: (0.0762) (40) [(0.254) (0.0762) ](3) (0.254) U4 4 4 \u3c0 \u3c0 \u3c0 + = + \u2212 = Solve for .Ans3U 6.33 m/s\u2248 3.37 A solid steel cylinder, 4.5 cm in diameter and 12 cm long, with a mass of 1500 grams, falls concentrically through a 5-cm-diameter vertical container filled with oil (SG = 0.89). Assuming the oil is incom- pressible, estimate the oil average velocity in the annular clearance between cylinder and container (a) relative to the container; and (b) relative to the cylinder. Solution: (a) The fixed CV shown is relative to the container, thus: 2 2 2 , : ( ) , . (a) 4 4cyl oil cyl oil Q Q or d V D d V thus Ans\u3c0 \u3c0= = \u2212 2 oil cyl2 2 dV V D d = \u2212 For the given dimensions (d = 4.5 cm and D = 5.0 cm), Voil = 4.26 Vcylinder. (b) If the CV moves with the cylinder we obtain, relative to the cylinder, = + = ( ) . (b)oil relative to cylinder part a cyl cylV V V V Ans 2 cyl2 2 D V 5.26 D d \u2248 \u2212 Chapter 3 \u2022 Integral Relations for a Control Volume 167 3.38 An incompressible fluid is squeezed between two disks by downward motion Vo of the upper disk. Assuming 1-dimensional radial outflow, find the velocity V(r). Solution: Let the CV enclose the disks and have an upper surface moving down at speed Vo. There is no inflow. Thus Fig. P3.38 2 out CV CS d dd V dA 0 ( r h) 2 rh V, dt dt \u3c1 \u3c5 \u3c1 \u3c1\u3c0 \u3c1 \u3c0 \ufffd \ufffd + = = +\ufffd \ufffd\ufffd \ufffd \ufffd \ufffd \ufffd \ufffd 2 o dh dh or: r 2rhV 0, but V (the disk velocity) dt dt + = = \u2212 As the disk spacing drops, h(t) \u2248 ho \u2212 Vot, the outlet velocity is V = Vor/(2h). Ans. 3.39 For the elbow duct in Fig. P3.39, SAE 30 oil at 20°C enters section 1 at 350 N/s, where the flow is laminar, and exits at section 2, where the flow is turbulent: 1/72 1 av,1 2 av,22 21 1 1r ru V u V RR \ufffd \ufffd \ufffd \ufffd \u2248 \u2212 \u2248 \u2212\ufffd \ufffd \ufffd \ufffd \ufffd \ufffd\ufffd \ufffd Fig. P3.39 Assuming steady incompressible flow, compute the force, and its direction, of the oil on the elbow due to momentum change only (no pressure change or friction effects) for (a) unit momentum-flux correction factors and (b) actual correction factors \u3b21 and \u3b22. Solution: For SAE 30 oil, \u3b3 = 8720 N/m3, thus Q = 350/8720 = 0.0401 m3/s. This flow Q must equal the integrated volume flow through each section: 2 2 av,1 av,2 ax,1 av,2 m mQ 0.0401 (0.05) V (0.03) V , or V 5.11 , V 14.2 s s \u3c0 \u3c0= = = = = Now apply the linear momentum relation to a CV enclosing the inlet and outlet, noting from Eqs. (3.43), p. 136, that \u3b21 \u2248 1.333 (laminar) and \u3b22 \u2248 1.020 (turbulent): 2 2 2 x 2 2 av,2 1 1 av,1 y 2 2 av,2F A V cos A V ; F A V sin\u3b2 \u3c1 \u3b8 \u3b2 \u3c1 \u3b2 \u3c1 \u3b8= \u2212 = (a) If we neglect the momentum-flux correction factors, \u3b2 \u2248 1.0, we obtain 2 2 2 2 xF (890) (0.03) (14.2) cos30 (890) (0.05) (5.11) 439 183 (a)Ans.\u3c0 \u3c0= ° \u2212 = \u2212 \u2248 256 N 168 Solutions Manual \u2022 Fluid Mechanics, Fifth Edition 2 2 yF (890) (0.03) (14.2) sin 30 . (a)Ans\u3c0= ° \u2248 254 N Whereas, if we include correction factors \u3b21 \u2248 1.333 and \u3b22 \u2248 1.020, we obtain Fx = (1.020)(439) \u2212 (1.333)(183) \u2248 205 N; Fy = (1.020)(254) \u2248 259 N Ans. (b) 3.40 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: x up up down down j j j j j j j F F m u m u m u m u , m A V\u3c1 \ufffd = \u2212 = + \u2212 = \u2212 = \ufffd \ufffd \ufffd \ufffd \ufffd Fig. P3.40 2 2 2 j jThus F A V (998) (0.05) (8) .Ans\u3c1 \u3c0= = \u2248 \u2190500 N 3.41 In Fig. P3.41 the vane turns the water jet completely around.