HW_11_solutions - 5.65 In turbulent flow near a flat...

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Unformatted text preview: 5.65 In turbulent flow near a flat wall, the local velocity u varies only with distance y from the wall, wall shear stress m», and fluid properties ,0 and M. The following data were taken in the University of Rléode Island wind tunnel for airflow, p = 0.0023 slug/ft), M = 3.81E—7 slug/(ft-s), and TW = 0.029 lbf/ft : 0.021 50.6 0.035 54.2 0.055 57.6 0.080 59.7 0.12 63.5 0.16 65.9 y, in u, ft/s (a) Plot these data in the form of dimensionless 11 versus dimensionless y, and suggest a suitable power-law curve fit. (b) Suppose that the tunnel speed is increased until u = 90 ft/s at y = 0.11 in. Estimate the new wall shear stress, in lbf/ftz. Solution: Given that u = fcn(y, w, p, M), then n = 5 and j = 3 (MLT), so we expect n — j = 5 — 3 = 2 pi groups, and they are traditionally chosen as follows (Chap. 6, Section 6.5): * 1=fcn pu y y 11* We may compute u* = (Tw 40)“2 = (0.029/00023)”2 = 3.55 ft/s and then modify the given data into dimensionless parameters: ), where u* =(rW/p)1/2 = the 'e'friction velocityl’ y, in: 0.021 0.035 0.055 0.080 0.12 0.16 pu*y/y: 38 63 98 143 214 286 u/u*: 14.3 15.3 16.2 16.8 17.9 18.6 When plotted on log-log paper as follows, they form nearly a straight line: 10° —-----Ill The slope of the line is 0.13 and its intercept (at yu */v = 1) is 8.9. Hence the formula: u/u* z 8.9(yu*/1/)0’13 : 1% Ans. (a) Now if the tunnel speed is increased until u = 90 ft/s at y = 0.11 in, we may substitute in: 0.13 9—0 z 8.9 = 8.9(54.5u*)0‘13, solve for u* z 4.89 ft/s u* 3.87E—7 Solve for 1W = pu*2 = (0.0023)(4.89)2 ~0.0551bf/ft2 Ans. (b) C/éggor . HWH fut/5; 6.28 For straightening and smoothing an Thousands airflow in a 50—cm—diameter duct, the duct ofstraws is packed with a “honeycomb” of thin straws of length 30 cm and diameter 4 mm, as in Fig. P628. The inlet flow is air at 110 kPa and 20°C, moving at an average velocity of 6 m/s. Estimate the pressure drop across the honeycomb. Solution: For air at 20°C, take H z 1.8E—5 kg/m's and p = 1.31 kg/m3. There . Fig. P628 would be approx1mately 12000 straws, but each one would see the average velocity of 6 m/s. Thus LV 32 . E— 0.3 .0 Aplaminar = = 8 )— z Pa AVIS. c12 (0.004)2 Check Re = de/‘u = (1.3l)(6.0)(0.004)/(1.8E—5) z 1750 OK, laminar flow. 6.52 The pipe flow in Fig. P652 is driven 3“ m by pressurized air in the tank. What gage Smomh pipe; Fv‘ r. Q pressure p1 is needed to provide a 20°C (i=5 Cm " ' "’ water flow rate Q = 60 m3/h? Solution: For water at 20°C, take p = 998 80 m kg/m3 and u = 0.001 kg/m-s. Get V, Re, f: 6 V = ——0/3600 2 = 8.49 2; (.777/4)(0.05) S Fig. P652 Re = 2, 424000; fsmooth z 0.0136 0.001 Write the energy equation between points (1) (the tank) and (2) (the open jet): 02 0 Vie 2 —p—1+—+10=——+ PP +80+hf, where hf=f£V— and Vie =8.49 2 pg 2g pg 2g d2g P" s 2 Solve p1 = (998)(9.81) 80 — 10 + (849) 1+ 0.0136 2(9.81) 0.05 ~2.38E6 Pa Ans. [This is a gage pressure (relative to the pressure surrounding the open jet.)] 6.62 Water at 20°C is to be pumped through 2000 ft of pipe from reservoir 1 to 2 at a rate of 3 ft3/s, as shown in Fig. P662. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient, What horsepower pump is needed? Solution: For water at 20°C, take p = 1.94 slug/ft3 and M = 2.09E—5 slug/ft's. For cast iron, take 8 -- 0.00085 ft, or s/d = 0.00085/(6/12) := 0.0017. Compute V, Re, Fig- P5152 and f. V=g=%=15.3 E; A (.71/4X6/12) s 1. 4 . R6 = p—Vd = w z 709000 8/d = 0.0017, fMood z 0.0227 g 2.09E—5 y The energy equation, with p1 = p2 and V1 z V2 z 0, yields an expression for pump head: 2 2 hPump = AZ +fEV— = 120 ft +0.0227 399—9 (153) = 120 +330 == 450 ft (1 2g 6/12 2(32.2) h 1. 4 2. . Power: P = fiQ—p = W = 112200 +550 z 204 hp Ans. 77 0.75 pm. Hw // fat/fad 6.102 A 70 percent efficient pump delivers water at 20°C from one reservoir to another 20 ft higher, as in Fig. P6102. The piping system consists of 60 ft of galvanized—iron 2—in pipe, a reentrant entrance, two screwed 90° long—radius elbows, a screwed—open gate valve, and a sharp exit. What is the input power required in horsepower with and without a 6° well— designed conical expansion added to the exit? The flow rate is 0.4 ft3/s. C(fcfié V 304/ Solution: For water at 20°C, take p = 1.94 slug/ft3 and M = 2.09E—5 slug/fts. For galvanized iron, 8 -- 0.0005 ft, whence e/d = 0.0005/(2/12 ft) »- 0.003. Without the 6° cone, the minor losses are: Kreentrant z Kelbows z I{gate valve z Ksharp exit z 1 Evaluate V=g=—9'4T=18.3 f—t; Re=p—W=W~2S4OOO A “(z/12) /4 s ,u 2.09E—5 At this Re and roughness ratio, we find from the Moody chart that f z 0.0266. Then 2 2 “0 hpump=AZ+V— f£+EK =20+(18'3) 0.0266 ~61 +1.0+O.82+0.16+1.0 2g d 202.2) 2/12 h 0r hpump e856 ft, P0wer= ng P = W 77 0.70 = 3052 + 550 z 5.55 hp Ans. (a) (b) If we replace the sharp exit by a 6° conical diffuser, from Fig. 6.23, Kexit z 0.3. Then 2 hp =20+ (183) 0.0266(fl) +1.0+ .82+.16+0.3 =81.95 ft 2(322) 2/12 then Power = (62.4)(0.4)(8 1 .95)/0 .7 + 550 e 531 hp (4% less) Ans. (b) ...
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HW_11_solutions - 5.65 In turbulent flow near a flat...

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