6.78 In Fig. P6.78 the connecting pipe is commercial steel 6 cm in diameter. Estimate the flow rate, in m 3 /h, if the fluid is water at 0 C. Which way is the flow? Solution: For water, take ρ = 998 kg/m 3 and µ = 0.001 kg/m s. For commercial steel, take ε 0.046 mm, hence ε/d = 0.046/60 0.000767. With p1, V1, and V all 0, the energy equation between surfaces (1) and () yields Fig. P6.78 p 00000 0 + 0 + z1 + 0 + z + h f, or hf = 15 5.43 m (flow to left) ρg 998(9.81) LV 50 V Guess turbulent flow: hf = f = f = 5.43, or: fv 0.178 dg 0.06(9.81) ε " 0.178# m = 0.00767, guess ffully rough 0.0184, V %.64, Re = 158000 d ' & 0.0184( s m fbetter 0.004, Vbetter =.50, Rebetter 149700, f3rd iteration 0.005 (converged) s The iteration converges to f 0.005, V.49 m/s, Q = (π/4)(0.06) (.49) = 0.00705 m 3 /s = 5 m 3 /h Ans. 1/
P6.94 Air at 0 C flows through a smooth duct of diameter 0 cm at an average velocity of 5 m/s. It then flows into a smooth square duct of side length a. Find the square duct size a for which the pressure drop per meter will be exactly the same as the circular duct? Solution: For air at 0 C and 1 atm, take ρ = 1.0 kg/m 3 and µ = 1.8E-5 kg/m-s. Compute the pressure drop in the circular duct: ρvd (1.)(5)(0.) ReD = = = 66, 700 ; fsmooth = 0.0196 µ 1.8E 5 3 1m 1. / L ρ kg m m Pa Δ p = f V = (0.0196)( )( )(5 ) = 1.47 D 0.m s m The square duct will have slightly different size, Reynolds number, and velocity: D h 4a ρvsquarea (1.) Vs a = = a ; ReDh = = 4a µ 1.8E 5 3 π m (0.) (5) 0.157 s π But Q = D V = = = V a 4 4 s Thus everything can be written in terms of the square duct size a: 1.(0.157 / a ) a 10470 Pa L ρ 1m 1. 0.157 Re Dh = = ; Δ p = 1.47 = f V = f ( )( ) 1.8 E 5 a m D a a 0.0147 f or : 1.47 = or : f = 99.5a 5 a 5 h s Guess f equal to, say, 0.0, find the improved Reynolds number and f, finally find a: Vs = 4.70 m/s ; Re Dh = 57,350 ; f = 0.003 ; a = 0.183 m Ans.
514 Solutions Manual Fluid Mechanics, Seventh Edition 6.109 In Fig. P6.109 there are 15 ft of -in pipe, 75 ft of 6-in pipe, and 150 ft of 3-in pipe, all cast iron. There are three 90 elbows and an open globe valve, all flanged. If the exit elevation is zero, what horsepower is extracted by the turbine when the flow rate is 0.16 ft 3 /s of water at 0 C? Fig. P6.109 Solution: For water at 0 C, take ρ = 1.94 slug/ft 3 and µ =.09E 5 slug/ft s. For cast iron, ε 0.00085 ft. The +, 6+, and 3+ pipes have, respectively, (a) L/d = 750, ε/d = 0.0051; (b) L/d = 150, ε/d = 0.0017; (c) L/d = 600, ε/d = 0.0034 The flow rate is known, so each velocity, Reynolds number, and f can be calculated: 0.16 ft 1.94(7.33)(/1) Va = = 7.33 ; Re a = = 113500, fa 0.0314 π(/1) /4 s.09e 5 Also, Vb = 0.8 ft/s, Reb = 37800, fc 0.066; Vc = 3.6, Rec = 75600, fc 0.087 Finally, the minor loss coefficients may be tabulated: sharp + entrance: K = 0.5; three + 90 elbows: K = 3(0.95) + sudden expansion: K 0.79; 3+ open globe valve: K 6.3 The turbine head equals the elevation difference minus losses and the exit velocity head: t = Δ f m c h z h h V /(g) (7.33) = 100 [0.0314(750) + 0.5 + 3(0.95) + 0.79] (3.) (0.8) (3.6) (0.066)(150) [0.087(600) + 6.3 + 1] 7.8 ft (3.) (3.) The resulting turbine power = ρgqht = (6.4)(0.16)(7.8) 550 1.3 hp. Ans.
7.9 Repeat the flat-plate momentum analysis of Sec. 7. by replacing Eq. (7.6) with the simple but unrealistic linear velocity profile suggested by Schlichting [1]: u y for 0 y U Compute momentum-integral estimates of c f, /x, */x, and H. Solution: Carry out the same integrations as Section 7.. Results are less accurate: u u y y u / (1 ) dy (1 ) dy ; * (1 ) dy ; H 3.0 U U 6 U / 6 0 0 0 U d d( /6) 1 3.64 w U U ; Integrate : dx dx x Re Re Substitute these results back for the following inaccurate estimates: 0.577 * 1.73 c f ; ; H 3.0 Ans. ( a, b, c, d) x Re x Re x x x x
P7.6 Consider laminar flow past the square-plate arrangements in the figure below. Compared to the drag of a single plate (1), how much larger is the drag of four plates together as in configurations (a) and (b)? Explain your results. Fig. P7.6 (a) Fig. P7.6 (b) Solution: The laminar formula CD 1.38/ReL 1/ means that CD L 1/. Thus: const (a) Fa (4 A1) 8 F1.83F1 Ans. (a) L 1 const (b) Fb (4 A1 ).0F1 Ans. (b) 4L 1 The plates near the trailing edge have less drag because their boundary layers are thicker and their wall shear stresses are less. These configurations do not quadruple the drag.
P7.35 Repeat Problem 7.6 for turbulent flow. Explain your results. Solution: The turbulent formula C 1/7 D 0.031/ReL m eans that CD L 1/7. Thus: const (a) Fa (4 A 1/7 1) 3.6F 1 Ans. (a) ( L ) 1 const (b) Fb (4 A 1/7 1) 3.8F 1 Ans. (b) (4 L ) 1 The trailing areas have slightly less shear stress, hence we are nearly quadrupling drag.
P7.53 From Table 7., the drag coefficient of a wide plate normal to a stream is approximately.0. Let the stream conditions be U and p. If the average pressure on the front of the plate is approximately equal to the free-stream stagnation pressure, what is the average pressure on the rear? Fig. P7.53 Solution: If the drag coefficient is.0, then our approximation is? Fdrag.0 U A plate (pstag p rear )A plate, or: prear pstag U Since, from Bernoulli, pstag p U, we obtain prear p U Ans.
P7.7 A settling tank for a municipal water supply is.5 m deep, and 0 C water flows through continuously at 35 cm/s. Estimate the minimum length of the tank which will ensure that all sediment (SG.55) will fall to the bottom for particle diameters greater than (a) 1 mm and (b) 100 m. Fig. P7.7
Solution: For water at 0 C, take 998 kg/m 3 and 0.001 kg/m s. The particles travel with the stream flow U 35 cm/s (no horizontal drag) and fall at speed Vf with drag equal to their net weight in water: 3 w 4(SG 1)gD W net (SG 1) wg D Drag CD Vf D, or: Vf 6 4 3C D where CD fcn(red) from Fig. 7.16b. Then L Uh/Vf where h.5 m. 4(.55 1)(9.81)(0.001) (a) D 1 mm: V f, iterate Fig. 7.16b to CD 1.0, 3C D (0.35)(.5) ReD 140, Vf 0.14 m/s, hence L Uh/V f 6.3 m Ans. (a) 0.14 4(.55 1)(9.81)(0.0001) (b) D 100 m: V f, iterate Fig. 7.16b to CD 36, 3C D 0.35(.5) ReD 0.75, Vf 0.0075 m/s, L 10 m Ans. (b) 0.0075
P7.75 The helium-filled balloon in Fig. P7.75 is tethered at 0 C and 1 atm with a string of negligible weight and drag. The diameter is 50 cm, and the balloon material weighs 0. N, not including the helium. The helium pressure is 10 kpa. Estimate the tilt angle if the airstream velocity U is (a) 5 m/s or (b) 0 m/s. Fig. P7.75
Solution: For air at 0 C and 1 atm, take 1. kg/m 3 and 1.8E 5 kg/m s. For 3 helium, R 077 J/kg K. The helium density (10000)/[077(93)] 0.197 kg/m. The balloon net buoyancy is independent of the flow velocity: 3 3 Bnet ( air He) g D (1. 0.197)(9.81) (0.5) 0.644 N 6 6 The net upward force is thus Fz (Bnet W) 0.644 0. 0.444 N. The balloon drag does depend upon velocity. At 5 m/s, we expect laminar flow: 1.(5)(0.5) (a) U 5 m : ReD 167000; Table 7.3: CD 0.47 s 1.8E 5 1. Drag CD U D 0.47 (5) (0.5) 1.384 N 4 4 1 Drag 1 1.384 Then a tan tan Ans. (a) F 7 0.444 z (b) At 0 m/s, Re 667000 (turbulent), Table 7.3: CD 0.: 1. 1 9.43 Drag 0. (0) (0.5) 9.43 N, b tan 87 Ans. (b) 4 0.444 These angles are too steep the balloon needs more buoyancy and/or less drag.