HW#10%20solution

# HW#10%20solution - 8.7 A soft drink with the properties of...

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Unformatted text preview: 8.7 A soft drink with the properties of 10 "C water is sucked through a 4-mm-diameter, 0‘25- m—long straw at a rate of 4 cm’ls. Is the ﬂow at the outlet of the straw laminar? Is it fully devel— oped? Explain. _ a -2._/_n_' v= 7% = = 0.319111 Tim, my» 1/=I.3I x/o"‘:’§??—z (seeﬁbleﬂﬂ m ~3 = M2: ﬂalgg’lmno m) __ I ' . . . =0.0¢€ Re , or Z; = (0.06DM7I )(W/o’sm) = 0.233 m < 0.25»: 7)- or we have fuI/y’dege/opea’l’ flow «7‘ {he 6in 8.16 ‘ M 8.16 Water at. 20! “C ﬂows through a horizon— tal 1-mm—diameter tube to which are attached two pressure taps it‘distance 1 m apart; (a) What is the maximum pressure drop allowed if the ﬂow is to be laminar? (b) Assume the manufacturing tolerance on the tube diameter is D '= 1.0 i 0.1 mm. Given this uncertainty in the tube diameter. what is the maximum pressure drop allowed if it - - ~06an must be assured that the ﬂow is laminar? Ram 8'2 V—lﬂOX/O S [LT/.001“; 3&3 a) War/Mum A}! corresponds 7‘0 maxi/WI» V) ‘or Re=¥70~=2ll00 ( w"; __ 2/001! _ 21,00 “[0 3—) 7771/3, V" r - .. m i D V Ib‘TrT-F _ 2'10? Fbr Mini/var How NI: weal. _ 32,421V_ iz(lxli3w),(zm)gio@ ' 32,”! I or AP” ﬁll?- ’ ‘ (léL-a'mf ’1 Thus, 4 AF = 6.711x/0 b) Since Vang/302!— cma' Ada-12%;!— I'l' fol/aw: f/mf 9 AW = 32/“Jbg7f’0l’ﬂ 77143;, 7776 Myer Me Jib/718W; 71/79 .sma/Ie'r Me 11f allowed *0 Mai/hfth kin/bar ﬂow. 777qu cpﬂsidﬁr D: ‘ /-/X/0“3/I7, 0" i 32. (/XA'O"; ) (lm )(Zlo‘o)(lx/6‘{—¢"—z) AW ( I.IX/0‘3m)3' = 5. 05 x/o“~n§'a 846 5 o 8:381——_+Wm 5.38 Water ﬂows at a rate of 10 gallons per minute in a new horizontal 0.75—in1.-dliameter galvanized iron pipe. Deter- mine the pressure gradient, Ap/(f, along the; pipe. a! Im'n 231m? ’4’ .. if \$103377 253'“) 7W)(77216W)’0'0223 5 Thus; v— % ( *ia'fi) Now] {or a borizom‘al pipe 32-»22'7éf 4,01: "fsz where 3/2206 v I 076’ , VD __. 7.27%(f'21’w‘) 4‘ ‘ ""”r"""‘ ., ‘ =3.76’X/0 Re V MIX/05 225—2 and é = 0.057053% .3 0‘008 D (9%51‘2‘) . if fol/aw: from 39.8.20 Mail {5 0.037 77103, ‘ IA IH‘ 0'6‘ TF3 ( I‘M/n.“ = 0. 27/1 psi/ff 9f; WNW“ = 3 gym (2) * 1 w J \ 8.65 Gaffes ﬂows thmugh mg ﬁlter shown in Fig? P8,..65 a; a rate 0! 4- tnugs/min == 60 ins/min. Dawning the coefﬁcient for the ﬁlter. Show that it is reasonable to neglect major losses. I FIGURE. P8.65 +2]: 777w) h4= Z, = if} If we neg/ed lame: ofﬁef' Ma” across f/ye {ﬁfe/a) {ﬂew 5L: K “V1 or = ‘fi/fer 7*; ’ ﬁ/ft‘r where ;,,3 ‘ HF Imfin (go-2- M" V3 3,22; = 0,0255%?! 9L 4 f m“?! m: zﬁ)(iﬁ)§ — ‘ ' .5" I1 3 _, 4 my!” " " (0'02 ~ 3.06x/0 ﬂ/m‘a’ Wm loss coefﬁbienis (enfmngcle, exit dc.) are mob/I less f/zan Mi: and can be neg/eded. 4M7 r I assemed incempxessible, flows through the we ’ pipes shown Fig. 198.97. Determine ﬁcwram iif mini)! losses are neglected and the friction facmr in each pipe is 0.015. Determine the ﬂowrate if the 0.5-in.—diameter pipe were re: placed by a Lin-diameter pipe. Comment on the assumption‘ of incompressibility. FIGURE 98.377 n 494 £132., =I7Ll +5“ + 3 ﬁg; 423 I tube/‘3 14:0}, zo=ZzJﬂq=Q (I) . 1 ‘ 3 “ A Dz. ’_ 0.5.1 3 \é=V,/zz,=£%f-¥; , g, and wwww— T/wsj [7.0) becomes =0.25\4 , _ 12 0.25Mz I V31 V1 42 - ﬁr: + as: + i; or , :1 Wm +11 ,5 ﬂ g / - . z :11 02 gm W’M 70” 9R7; or ‘00 (/7/6sﬁ'lé§)(lsowwo)"ﬂ 00 m H: and f,=fz= 0.015 (1) gives 7 2b , ling“ _ sl 2 3,. 2:0}? 2 20H (0.5 EQUW—ﬁi) uf (0.002679%)‘4 [(0,015 [Jig )(o.2.5)+ if} or 14: 70.4 {gt 7%, 42:”: V = f(-2-,’zﬁ)2/‘ia‘f§5) =o./23 I; If 507% pipes Ware //'0. c/I'a_/;v7.ef¢mr“J 2%.?» W ‘44 am, (I) beck/21:5 ff %e\é’[£ 300 =ZPKZEQDJ§ H] Hence) (0.5 %)(/4¥£€ ) = {10.0on ) szlawq DH] H] or MW} £379., 4‘12, 400’ 0,50; 405‘ 754- H] 0” \ 3 14 =9/.7 # ms, 0 =/2z 14 ~%(,—;—H)’(9/.7§t) = 0.500% Since (70 = PR7— i7l fol/am ffhm“ ‘ £3. 3 , —%: = ﬂ = 5% [fa/e aux/me 7} =75 (IV/Omiaé/y 1017/00/59, 1502’ I'l’séou/a’ be a remomb/e aﬁpjpm‘maf/m) 7%” e? A, , /6‘.7 «I' F N j _ —_£L_ (0.5 mews/1 " 0'75 7 . 7776’ WW near/I Mamas/We» 8’46 ﬂame The ﬂowrate between tank A and tank showu in Fig. P8.100 is to be increased by 330% (Le. from Q to 1.30Q) by the addition of sec» and pipe (indicated by the dotted lines) rulinng from node C to tank B. If the elevation of the: free surface in tank A is 25 ft above that in tank r 3'"- B . determine the diameter, D, of this new ‘ipe. """W’R'ES'" Neglect minor losses and assume that the frietion factor for each pipe is 0.02. ‘ 6-in.diam,etet; 6—in. diameter; 600 ft tong 590 ft long Diametet D. 500 ft long FIGURE P8400 i . ' ‘ : g a _ V2 a a Mil/7 f/Ie Slay/ep/pe é§+g+zﬂ =: @213; +zg+ﬂ .gfil/ﬁ Where ﬂy:ﬂ8=01 %=l/g%0’z,9 =25f7’l 25 so, and V, = K (sf/ice D, =0; ). 2 .. Id) 3 = (1°2’*5°09ff __..._____VI _ 77m, 2,, — {it}. l or 25 f/ (0.02) (ﬁg) 26321 g) or W = 6405 g Hence} 0:141, V, =f(f§ff)2(6.05 £1) : Ales I 3 Vii/1 4% second pipe «9-3 It amazes £5) = Asa-‘2'— T/ws, 051.54% =02 +69% or 14: 73;. = = 7.84133 Fér fluid flowing from K) foéB iii/wig}; p/pgs I and 2 I = I. ,2 g 1” V“ . 2/9 zbt/ + 12 a; {£2 '3: 7;- (599 [fl/J) or r; (7wi £3)“ 5 H v1 252‘! = (0.02 50° + 0.02) 09w . '_2__ ) (765.545 432139;) ( (7-3; H)— 2632.2 £11) and 3 J . Q1: ’91 V2 = 516% {071605;} 20. 51/ 777%" 93 = 0: " 02. = L\$¢§J—0.Sllg—s = [’03ng Fir fluid ﬂowing from )9 I08 wrong/i pipes / a043, 3 zﬂ=bq+b£ = Alzﬁﬂ-é-Y; 1.03%“; 1.31 633 I D’ 7? 3 '03 2.; Rife/~43 “3:7- :1 ‘1 _ : ﬂ 7771/5; ' 1 J 3 4 D, D i 1.31 2 = . 600 500 ; 25!! (0.02) (7%”) 2-Wqu No.02) ———ﬂ—D3 LLzélzg) K OI“ law-6m Nole -' Wi/A Me perv/ne’er: Earl/en, Me 30/1/7901; 11: 70/79 sew/7W0 1’0 rot/m! 019' «mm in 7’60 #ﬂ/CVIJfI-‘WJ ...
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## This note was uploaded on 05/04/2010 for the course TAM 335 taught by Professor Whoever during the Spring '08 term at University of Illinois at Urbana–Champaign.

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HW#10%20solution - 8.7 A soft drink with the properties of...

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