P3.18 An incompressible fluid flows steadily through the rectangular duct in the figure. The exit velocity profile is given by u umax(1 – y2/b2)(1 – z2/h2). (a) Does this profile satisfy the correct boundary conditions for viscous fluid flow? (b) Find an analytical expression for the volume flow Q at the exit. (c) If the inlet flow is 300 ft3/min, estimate umax in m/s. Solution: (a) The fluid should not slip at any of the duct surfaces, which are defined by y b and z h. From our formula, we see u 0 at all duct surfaces, OK. Ans. (a) (b) The exit volume flow Q is defined by the integral of u over the exit plane area: 22 2 2max max max22 2 24411 1 133hb b hhb b hyzyzbhQ u dA u dy dz u dy dz ubh b h . ( )bAnsmax16bhu9 (c) Given Q 300 ft3/min 0.1416 m3/s and b = h = 10 cm, the maximum exit velocity is 3maxm160.1416 (0.1 m)(0.1 m) , . (c)s9Qusolve forAnsmaxu7.96 m/s 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 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. P3.33Area of test section = 22 2 (0.4 )(4 ) 10.053rL m m m The number of holes on the test section is 21 :1200 10.0531: , 12064mNN 23suction holeQ NQ NAV (12064)( /4)(0.005 m) (8 m/s) 1.895 m /sr 22oo1 oo(a) Find V : Q Q or V (2.5) (35) (0.8) ,44solve for V . (a)Ansm3.58s 2221suction 22(b) Q Q Q (35) (0.8) 1.895 V (0.8) ,44or: V . (b)Ans m31.2s 22ff2 ff(c) Find V : Q Q or V (2.2) (31.2) (0.8) ,44solve for V . (c)Ansm4.13s P3.49 The horizontal nozzle in Fig. P3.49 has D1 12 in, D2 6 in, with p1 38 psia and V2 56 ft/s. For water at 20°C, find the force provided by the flange bolts to hold the nozzle fixed. Solution: For an open jet, p2 pa 15 psia. Subtract pa everywhere so the only nonzero pressure is p1 38 15 23 psig. The mass balance yields the inlet velocity: 11 2 2VA =VA , 2211ftV (12) (56) (6) , V 14 44 s The density of water is 1.94 slugs per cubic foot. Then the horizontal force balance is 22 11F=mu-muX 2x bolts 2 2 1 1 2 1F F (23 psig) (12 in) m u m u m(V V )4 2boltsft ftCompute F 2601 (1.94) (1 ft) 14 56 14 .4ssAns 1700 lbf P3.61 A 20°C water jet strikes a vane on a tank with frictionless wheels, as shown. The jet turns and falls into the tank without spilling. If 30°, estimate the horizontal force F needed to hold the tank stationary. Solution: The CV surrounds the tank and wheels and cuts through the jet, as shown. We should assume that the splashing into the tank does not increase the x-momentum of the water in the tank. Then we can write the CV horizontal force relation: xininjettankdF F u d m u 0 mV independent of dt 222jj3slug 2 ftThus F A V 1.94 ft 50 .412 sftAns 106 lbf Fig. P3.61P3.95 A cylindrical water tank discharges through a well-rounded orifice to hit a plate, as in Fig. P3.95. Use the Torricelli formula of Prob. P3.81 to estimate the exit velocity. (a) If, at this instant, the force F required to hold the plate is 40 N, what is the depth h ? (b) If the tank surface is dropping at the rate of 5 cm every 2 seconds, what is the tank diameter D? Solution: For water take = 998 kg/m3. The control volume surrounds the plate and yields (b) In 2 seconds, h drops from 1.63m to 1.58m, not much change. So, instead of a laborious calculus solution, find Qjet,av for an average depth hav = (1.63+1.58)/2 = 1.605 m: P3.131 In Fig. P3.131 both fluids are at 20°C. If V1 1.7 ft/s and losses are neglected, what should the manometer reading h ft be? Solution: By continuity, establish V2: h d = 4 cmFD CV).()/81.9)(2()04.0)(4/)(/998(40:)2()4/(Thus;2saysTorricelliBut4)()(2232222aAnssmmmkgNhdataGivengdFhghVVdVVAVmumFFjetjetjetjetjetjetjetininxm1.63).()05.0)(4/()2)(00705.0()4/(:or,Equate/00705.0)605.1)(/81.9(2)04.0(42tank322bAnsmshtQDhAtQsmmsmmghAQavjetavm0.60Fig. P3.9511 2 2AV AV ,222112ftV V (D /D ) 1.7(3/1) 15.3 s Now apply Bernoulli between 1 and 2 to establish the pressure at section 2: 22111222p V gz p V gz ,22 Fig. P3.131 221 1or: p (1.94/2)(1.7) 0 0 (1.94/2)(15.3) (62.4)(10), p 848 psf This is gage pressure. Now the manometer reads gage pressure, so 1a merc water2lbfp p 848 ( )gh (846 62.4)h, solve for h .ftAns 1.08 ft P3.137 In Fig. P3.137 the piston drives water at 20°C. Neglecting losses, estimate the exit velocity V2 ft/s. If D2 is further constricted, what is the maximum possible value of V2? Fig. P3.137 Solution: Find p1 from a freebody of the piston: xa111 1a221F 10.0 lbf lbfF F p A p A , or: p p 28.65 A( /4)(8/12) ft Now apply continuity and Bernoulli from 1 to 2:2212a111 22 1 2VVp1pVA VA , or V V;422 2222(28.65)V,V5.61.1.94(1 1/16)Ansfts If we reduce section 2 to a pinhole, V2 will drop off slowly until V1 vanishes: 22(28.65)Severely constricted section 2: V .1.94(1 0)Ansft5.43s P3.155 The centrifugal pump of Fig. P3.155 has a flow rate Q and exits the impeller at an angle 2 relative to the blades, as shown. The fluid enters axially at section 1. Assuming incompressible flow at shaft angular velocity , derive a formula for the power P required to drive the impeller. Solution: Relative to the blade, the fluid exits at velocity Vrel,2 tangent to the blade, as shown in Fig. P3.116. But the Euler turbine formula, Ans. (a) from Example 3.18 of the text, 2t2 1t12t2 t1Torque T Q(r V r V )Qr V (assuming V 0) involves the
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