HW 4 Solution

HW 4 Solution - 6.78 In Fig P6.78 the connecting pipe is...

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Unformatted text preview: 6.78 In Fig. P6.78 the connecting pipe is commercial steel 6 cm in diameter. Estimate the ﬂow rate, in mS/h, if the ﬂuid is water at 20°C. Which way is the ﬂow? Solution: For water, take p = 998 kg/m3 and ﬂ = 0.001 kg/m-s. For commercial steel, take 8 2 0.046 mm, hence a’d = 0.046/60 = 0.000767. With p1, V1, and V2 L=5°m all z 0, the energy equation between Fig. P6.78 surfaces (1) and (2) yields 0+0+zl z Bi+0+zz +hf, or hf =15—M'z —5.43 m (ﬂow to left) (— Pg 998(9.81) 2 2 Guess turbulent ﬂow: hf = #111431 V = 5.43, ‘or: W2 = 0.1278 (1 2g 0.06 2(9.8_1) 1/2 3.32;) 2264 2, Re=158000 ’3: 0.00767, guess fwny mug, = 0.0184, V =( S fm, z 0.0204, Vbetter = 2.50 31, Rebetter z 149700, gmnmm 4 0.0205 (converged) S The iteration converges to f = 0.0205, V 2 2.49 m/s, Q = (n74)(0.06)2(2.49) = 0.00705 111% = 25 111% <— Ans. 6.83 For the system of Fig. P655, let Az = 80 m and L = 185 m of cast-iron pipe. What is the pipe diatneter for which the ﬂow rate will be 7 m3/h? Solution: For water, take p = 998 kg/m3 and y = 0.001 kg/m-s. For cast iron, take 82 0.26 mm, but (1 is unknown. The energy equation is simply Fig. P5-55 8fLQ2 _ 8f(185)(7/3600)2 _ 5.78E—5f 77:2gd5 752(9.81)d5 d5 A2 = 80 m = hf = , or d e 0.0591f”5 4 Guess f z 0.03, d = 0.0591(003)”5 e 0.0293 In, Re = 75% z 84300, 5: 0.00887 Iterate: fbener .~. 0.0372, dbme, z 0.0306 m, Reba“er z 80700, ddlbem z 0.00850, etc. The process converges to f z 0.0367, (1 z 0.0305 In. Ans. 6.101 In Fig. P6.101 a thick ﬁlter is being tested for losses. The ﬂow rate in the pipe is 7 m3/min, and the upstream pressure is 120 kPa. The ﬂuid is air at 20°C. Using the water- manometer reading, estimate the loss coefﬁcient K of the ﬁlter. Fig. P6.101 Solution: The upstream density is pair 2 p/(RT) = l20000/[287(293)] = 1.43 kg/m3. The average velocity V (which is used to correlate loss coefﬁcient) follows from the ﬂow rate: 3 Q =_7£0_"_1_/S_2 = 14.85 m/s AW. (7r/4)(0.1 m) The manometer measures the pressure drop across the filter: V: Apmm = (pw — pa)ghmam = (998 —1.43 kg/m3)(9.81 m/s2)(0.04 m) = 391 Pa This pressure is correlated as a loss coefﬁcient using Eq. (6.78): Apﬁh‘cr 39] Pa Kﬁlter = 2 = 3 2 z 2.5 ‘ (l/2)pV . (1/2)(1.43 kg/m )(14.85 m/s) Ans. 6.121 Consider the three—reservoir system of Fig. P6.121 with the following data: [41295111 [422125111 L3=160m z,=25m z2=115m 23:85m All pipes are 28-cm—diameter unfinished concrete (8 = 1 mm). Compute the steady ﬂow rate in all pipes for water at 20°C. Fig. p6_121 Solution: For water at 20°C, take p = 998. kg/m3 and ,u = 0.001 kg/m-s. All pipes have dd 2 1/280 = 0.00357. Let the intersection be “a.” The head loss at “a” is desired: 2 2 .. . . 2 21—ha:flldl\2]l; 22—ha=f2£‘l—Zz—; Zs‘ha zifsﬁ‘vi 1 g d2 28 d3 2g plus the requirement that Q + Q2 + Q3 = 0 or, for same d, V1 + V2 + V3 = 0 We guess ha then iterate each friction factor to ﬁnd V and Q and then check if 2Q = 0. 2 h3 = 75 m: 25—75 = (—)50 = f, (—95—) V1 , solve f1 = 0.02754, V1 z -10.25 —“—‘ 0.28 2(9.81) 8 Similarly, 115 —75 = f2(125/0.28)[v§/2(9.81)] gives f2 z 0.02755. v2 2: +7.99 and 85 — 75 = f3(160/0.28)[V32/2(9.81)] gives f3 z 0.02762, V3 z +3.53 1:1, 2V =+1.27 Repeating for ha = 80 m gives V1 = —lO.75, V2 2 +7.47, V3 2 +2.49 m/s, 2V = —0.79. Interpolate to ha z 78 m, gives V1 = —lO.55 m/s, V2 2 +7.68 m/s, V3 = +2.95 m/s, or: Q1 = —0.65 m3/s, Q2 = +0.47 m3/s, Q3 =‘ +0.18 m3/s. Ans. ﬂare m élMev-wa. SO‘F So‘V'W‘ 4AA- M well- 7 j , yaw Aave ~quee 23145, gull—i1 #m [4,2 Flats ﬁe Sam 6‘? 4L; 0’2 2728: 76m. He A.” 77.004222 +76 Awe madam 0’; “(14¢ 4th Vdewiaie. (Lyrics: your {afﬁrmiwwn C4" M “HN‘ SOIJW‘) 4145 is M6557 .lpgolve‘ HawCVeV; Lil-f; PossiL/e 796/1741)th Varialoiaf 49 7d" cHae 52,114ka ...
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