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 Q u dA h b h b b y2 z2 y2 h z2 4b 4h umax 1 2 1 2 dy dz umax 1 2 dy 1 2 dz umax b h h 3 3 b h b 16bhu max 9 Ans b c Given Q 300 ft3 min 0 1416 m3 s and b h 10 cm the maximum exit velocity is Q 0 1416 m 3 16 0 1 m 0 1 m umax solve for u max 7 96 m s s 9 Ans c 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 33 Area of test section 2 rL 2 0 4m 4m 10 053m 2 The number of holes on the test section is 1m 2 1200 10 0531 N N 12064 Qsuction NQ hole NAVr 12064 4 0 005 m 2 8 m s 1 895 m3 s a Find Vo Q o Q1 or Vo 4 m solve for Vo 3 58 s b Q 2 Q1 Qsuction 35 4 or Vf 4 0 8 2 Ans a 0 8 2 1 895 V2 m or V2 31 2 s c Find Vf Q f Q 2 2 5 2 35 4 0 8 2 Ans b 2 2 2 31 2 4 m solve for Vf 4 13 s 4 0 8 2 Ans c 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 V1A1 V2 A 2 V1 4 12 2 56 4 6 2 V1 14 ft s The density of water is 1 94 slugs per cubic foot Then the horizontal force balance is FX m 2 u 2 m 1u1 Fx Fbolts 23 psig 4 2 u2 m 1u1 m V 2 V1 12 in 2 m Compute Fbolts 2601 1 94 ft ft 1 ft 2 14 56 14 1700 lbf 4 s s Ans 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 d in u in 0 mV jet independent of Fx F m u d tank dt 2 2 slug 2 ft Thus F A jV 2j 1 94 3 ft 50 106 lbf Ans s ft 4 12 Fig P3 61 P3 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 CV to estimate the exit velocity a If at this h instant the force F required to hold the F plate is 40 N what is the depth h d 4 cm b If the tank surface is dropping at the D rate of 5 cm every 2 seconds what is the tank diameter D Fig P3 95 Solution For water take 998 kg m3 The control volume surrounds the plate and yields Fx F m in u in m jet V jet A jetV jet V jet But Torricelli says Given data h 2 V jet 2 gh Thus h 4 2 d 2V jet F 4 d 2 2 g 40 N 998kg m 4 0 04m 2 2 9 81m s 2 3 1 63 m Ans a 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 Qav A jet 2 ghav 4 0 04m 2 2 9 81m s 2 1 605m 0 00705 m 3 s Equate Q t Atank h or D Q t 4 h 0 00705 2s 0 60 m Ans b 4 0 05m 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 A1V1 A 2 V2 V2 V1 D1 D2 2 1 7 3 1 2 15 3 ft s Now apply Bernoulli between 1 and 2 to establish the pressure at section 2 p1 2 V 21 gz1 p2 2 V 22 gz 2 Fig P3 131 or p1 1 94 2 1 7 2 0 0 1 94 2 15 3 2 62 4 10 p1 848 psf This is gage pressure Now the manometer reads gage pressure so p1 pa 848 lbf merc water gh 846 62 4 h solve for h 1 08 ft ft 2 Ans 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 Fx F pa A1 p1A1 or p1 pa Now apply continuity and Bernoulli from 1 to 2 F 10 0 lbf lbf 28 65 2 2 A1 4 8 12 ft V1A1 V2 A 2 or V1 V 22 1 V2 4 p1 V 21 2 28 65 ft V2 5 61 s 1 94 1 1 16 2 pa V 22 2 Ans If we reduce section 2 to a pinhole V2 will drop off …
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