hmwk 9 solutions

hmwk 9 solutions - i 8.93 Water is circulated from a large...

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Unformatted text preview: i 8.93 Water is circulated from a large tank, through a filler," d hack to the tank as shown in Fig. P893. The power added to the water by the pump is 260 ft- “3/5. Belem-Linc the flowlpile I through the filter. \200 f1.nHJ,1-fl-din. ‘ pipe with em = 0.01 E FIGUHE P8311 ; k i v‘ V1 %+z,+i-a—q'-+-hp=4f£+z +5; D wbere fizp, J V, =V1=o, and fir-‘32. Also, L54: ramp or h .—. 422$: m P 52.#fi%{,§m/HJ‘)Vi v 77mg Eq, (I), became: . 2 409 2001»? ! V ._v._ = (mf +(0.8+53(l.5) +12+6 +0) M31132) "3 _ 13.13 i v z ——-—— § (f+0.01355) 1 7M1! and error salt/I‘M)“ fissume f= 0.04 .. Ham} Ey. (2), V= 6.26 5 from ff. ('3), Re = 5,2ox/a“. Tfim; Frag 5'9. 9.25; f: o. 039 -'# 0.04: 193502179 PIMP?) or V: 6-i29'3fl and We =522/X/04afld 10:19.03? (Checks) 7%ka [Q =I9V== £(0.IHJ1(F.29§5)= 0.04:9??? HHernafit/eéi, #75 Ca/gerok much“ (5?, 835) could be wed rafher Man {he Moody chin/"f. 77mg (aim ’1‘) ‘3—8? M where from 139;,(2;J (4:) f=(/3.13/|/3)—0.01365' ' (5) 77505; by cambinfng 13$. {3; [fliand-(SJ We obfain 1776* following Eyym‘z‘m for V3 l l /[( 13.13/v3) ~0-01365]4‘=—2. Ia Egg2.51/[ezfoanaJa/v3yo.anfiffi} Using a smut/fer mfi—fthdmy pro mm gills: {he solfiian 1‘0 EfJJ) as m __ H - V— 6.295— , Me same a: 0571mm 1 by Hm 4501/9 {rig} and eflw- mko i 8.97 Air. assumed incompressible, flows through the: was pipes shown in Fig. P837. Determine lhe flowratc if r‘n'mor' losses are neglected and um friction factor in each pipe is 01.016. Determine the Howrah: if the 0.5-in.-diume1r:r pipe waft re- placed by a MIL-diameter pipe. Comment on Lhe assumlpliunl of incumpmssibiiily. I FIGURE P837 +23} where %=0120=z;1fl3501 (I) ‘ 1- ‘ 2 . - 9- %,and|4=14%3=m%)= v2 777st £?_(/) becomes =6L2.‘51{;l _ 0.25V." 17 W a | or 2,0,, = £91611}; {2; (0.25324 {2% + I] (2) L511 ' {71.1 WI 100%“? or Po: 0 75 *(1-7/6 rug"? )(ISomofiR and fi= )9: 0.015 Eq. (2-) gives 5 I” R = 0.0020? Elf-‘3'? (0.5%,).(IW-gif) =19 (0.002 9%)fflo-05XfigN0-Ef fffl D” 1] 2+9 or 1/1: 970.4 gt 77m: Q: 1 t4 = f(i%'f{)2(9a#§)=o.123 g If 507% pipes Were / 1}). Jib/Dyer; 2%” V, =Vz and 0.} beam/71.9: gag gel/3R +76%: +1] 011" vii/1 Hz, 44:} m: 4:9; 2192) ‘ 2% = 2 P WE (*5;- “1 #801381 _J- 4 fit 11. .111.- =J_ 5:" 1 . 0 (0.537304%) 262-00209 )Ié[0°’5( TWP] 0r V1=9L7 gt ms, Q=flzm%%(fi—H)2(w.7§t)= 0.500%; 5/3059 69=FRT f7! Ital/aw £11611! 1 {Jo—a) If We «same 73 =75 (I'l’flNéqfi/y “HY/#01156", RTn ‘ 503' BMW/J be a remand/e oppkxm‘m’im) M917 / . ‘ . '6; x j = W} =0.§’67 _ 730 Mavis gawk Mchresslfi/e. | Bull?!) The fluwmte between tank A and itank 3 Shawn in Fig. 138.100 is to be increased by 30% _ 5-in-diamfiflzn 6-in- diameter: (i.e.. from Q to 1.3(3Q) by the addition of 'sec. - . . 5°” " "’"B 5"“ " “‘"E and pipe (indicated by the dotted lines) ru IHing, = from node C to tank B. If the elevation of the free surface in tank A is 25 ft above that in tank 3, determine the diamcter. D, of this new ipa. Neglect minor tosses and assume that the fri lion Nathan-‘1?- 509 It l'unB factor for each pipe is 0.02. a] FIGURE P8109 Mil/J file sing/e Pipe = where flfflgflaJ M}: and V, = K. (szhce D,‘ ). TM, 39 = 2‘, (11¢ it“; , or 25340.on (daemon) H V," ' (£85 2(32.2 1:1,) r i 3 " v, = 6.05 ti 0w =£G2Hfrm$ = we ii WM 4% Second pipe 0;} 1.300.193 ) = [-5.55 £15 x 3 77m, QI=L54¢~§E=QZ+Q$ or 1kg} =i£§$ =Z'ggzgt 4 75 For fluid flowing from 24 {0:8 threw/J pipe: land 2" - _ 1 1 g = 2,9 =51, 1%; 157$ 2% “‘92 £3: 3% {593 Eff”) 0!" a. H: . 600G rumba _ nary v,L 25 (092) GER, 432.2%) (0 02‘) (fiftj Haggai) Heme: V2: 2.60% I Grid :1 0;: fl; t4 = {56% It); (2-606?) =0-5H £5 77”" 623 = 0. d 01" 1.5410,”? = I. as? ' t For 1%?! flowing from H {eff threw/2 prpes I mam, H zfl=bir+b£3 = 1&5? 2;,wéere lé=£f=40317 "3' Tilly's: I ( 2 £0: IF = man (299?): M _&a__‘_‘ 25” I’D-“2’ (#9275312 1) “0'0” 03 292.2%) or DJ :7 O. 6621‘} Mule -' MW) 1%? parameters aim-1:, fie :a/m’rbn :1: gaffe semifim 2’0 rounder? arr-an: 1:1) #15 blew/MM; Elevation c: 60 m 8.102 The three water-filled tanks shown in Fig. P8.102 are connected by pipes as indicatecg. If minor losses are neglected. determine the flow- rate in each pipe. 1 _ 7 FIGURE P8492 ASSUME Hie fluid flows Win? )9 2‘03 and/41596. Mus, Q, =Qz+93 gyfmm)?‘ V, = —§(o.oam) y; + £(mam-fwg ’ 14 = 0,64 V2 +0.54: 14 m For fluid flowing From fife? wH/J flq=fl3=0 and Hq=lé=0J .2 2 39:38 +££LJ +££§ V1 or D, 2‘9 D: a? _ __ 200/): I V1" 20cm; V: 60m 20m - (0'0153( mm awafg) +(0.029)(m)m Hence, 5 5:0 : /.529 Vf + 2.55 v: i (2) Sim/affix, for f/Iéid flowing from x? 7‘06 .WIWJ £970; =0 and K942 =19J 24:26 + ‘23 £1 60m '= (0.0I5)(203;m 2m 1 f1) +(o.ozo)(3%3§£) zmfifif Hence,- I 50:].529 1/,2 +5Jovaz (3) Solve 59.5. (04(2), and (3) for V“ Va, Ma'Lé, Frw» @311) and (a): 50=L529(0.H)1(I/2+b3)2+449 [all or 95.8 =(V2H/3)‘+ 611‘sz (49 Subzlr‘acf £911) From 5?, (3) ;‘= 50-40 = 5.10 I42+ 2.551411'0r- 1/2 =W (5) . 1 2 7720.53. from 59¢. (We/761(5). EBJIJCIGIW f2V3z_7.5¥' +16) #4158 =0 t = 777/19 can be Sim/1'10}?! 1‘01 2 waif—284* = 10336 film/31 Square bail: side: and m rearrange 1‘0 give ‘4“ #9163 W2 + 42-5 =0 w/Jich am he SOIL/ea] by #79 quadrafic fanny/é fa nil/e 2 .SBiJ—ffi H? = M 1953 40:25) :H'Wor 7.86 Thus V3 2. | or V3=2.80gl I _ (gen?) (can’t) M’fe ‘ The Value lé=3"*$5fl [-9 1'70?l (1 so/Ufian of H18 original Eyum'r’afls, Efs. (I), (2), and (3). i Wflfi Mg Valve H73 rig/51' banal 5:21.? a)“ Efféj is fleqm'iua (:25. [0.3% —I!./’/- M32": 103.641.}? (3.413325‘2495)‘ ,9; Seen {ram Me [229’ Mm! side of £1716), “W: can/mt be. 7771': exfra r004 was t'nl'r‘aduced Squaring £716). Tm, a; = as 1/3 = 7202.0 mflmogl) = amt/£13 Also, from Eq. (331 ' 60:1.529 Mz+5i0 (2 so): a, 14: 3.52 or 3 and From 557.0): 3.52 = 0.54 14 +0.6Ifi(2.ea) or V2: 2.85% 01" l 3 Q1 =14; V: = §(o.aam)2(2.36 %) = 0. ants—g“- 8-103 ...
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This note was uploaded on 05/04/2008 for the course MMAE 310 taught by Professor Wark during the Spring '06 term at Illinois Tech.

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hmwk 9 solutions - i 8.93 Water is circulated from a large...

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