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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/(fts), 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
powerlaw 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 loglog 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/ms. 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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 Fall '08
 Schwartz,L

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