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set10-answers - 7 1"“ 7 , ., g i t E ,, i _ at...

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Unformatted text preview: 7 1"“ 7 , ., g i t E ,, i _ at ,r ‘ r v s k—b—d 7.14 As shown in Fig. P7.l4 and Video V5.4, a jet of liq- uid directed against a block can tip over the block. Assume that the velocity, V, needed to tip over the block is a function of the fluid density, p, the diameter of the jet, 1). the weight of the block, “W, the width of the block, b, and the distance, d. be- tween the jet and the bottom of the block. (a) Determine a set » of dimensionless parameters for this problem. Form the di— .7 mensionless parameters by inspection. (b) Use the momentum equation to determine an equation for V in terms of the other I r ' variables. (c) Compare the results of parts (a) and (b). ‘ {sieves 97.14 i i‘ w E , ‘ 1 I PI. .‘ ‘ a * ~ ” v. I r _ , ,,. f,‘_, i I . 2 _, i a V w _‘ € € 1 i i us; a t :, ' H i“ w ; , _ a . t 1 . . _ ,3 t 3 1 , _ 3: 3 g . _ , i , ; _ a a _ i 1 _ ,_ , I ., i. . a i (12?: " :lc'égr Impendwi‘vL/epmi army?th a: _ , : so? i i ,, l7 _ _ , 1 mlanmév £71m “.c‘anisx'a’ekafizééér 7W2" s ' r ‘ u i 1" ,. 1 i * i ‘ l : ' v 2 , ' , ' f V'I‘.§$140Wh' L s ‘ , ‘ é : . , , g ,. V, . ; , . i ‘ ;‘ , 5 . I K . L , . ; . V t t , , " _ a . _ ‘ . a . - - I ., , fi _ V " ‘ g 5 X E p ? 1 . , 1 ‘ ,, , Dz ‘ - = i i “ 5 i ‘ ? a; ' ‘ ‘ I _,J 1 H i i > 3 I r ’F;2:.i% A cylinder with a diameter, D, floats upright in a liquid as shown in Fig. P7 .20. When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium po- w; sition with a frequency, w. Assume that this frequency is a func- " tion of the diameter, D, the mass of the cylinder, m, and the specific weight, 'y, of the liquid. Determine, with the aid of dimensional analysis, how the frequency is related to these var- iables. If the mass of the cylinder were increased, would the frequency increase or decrease? : «m: , Lu; T“! Mr: FA’ITZ b’: Cylinder diameter = I) ' Chen/C “Slhj MLT: / ‘1) £9. YT, _; LT .61. -_—' yap/"D ,: oz D ‘2? * (L) MWT“ Shire. 77w: B 011/9 / pf 4-?le H‘ ice/lows 77ml- wkere C Is 4.. cansv‘un'LL'. 777‘“) Fran: 7713 “SR/15 1'7‘ AM:an 771075 [F rm :3 I'ncréaSed m 60H] decrease. o—.——o—-—__ Q. 7.29" | i 7.2!? A fluid flows through the horizontal curved pipe of Fig. P72? with a velocity V. The pressure drop, Ap, between the entrance and the exit to the bend is thought to beia function of the velocity, bend radius, R, pipe diameter, D, and fluid den- ..§,i.tzw 9- I139. datajhgwn, igflzgfpllowipg table were (themes in. the laboratory. For these tests p = 2.0 slugs/f9, R ‘=‘ 0.5 ft, and D = 0.1 ft. Perform a dimensional analysis andbased on the data given, determine if the variables used for this problem appear to be correct. Explain h0w you arrived at your answer. v (ft/s), 2.1 3.0 3.9 5.1 & Ap(ib/ft2) 1.2 1.8 6.0 6.5 I FIGURE 97.2? AP -' f/V, R, 1.3/0) 4/95 F4“; [/z-LLT" 2:31. 0:51. fli' FL—lfrz Fl’rym 777a )sz flea/em) 5— 3 .2 Z. [oi 246/7775 raga/krtt _ [3’7 M's/$15464, 1§r '77; (wnéelb/hj 4? )3 AP - (FL‘Z) h. — Chad/4 “5/47 x'L/LT Syskm 3 AF . (Ml—"l T—z) _;_ [(10 a 0 : "j ’W'" “‘ Z T pk /0 V" j (ML‘U (47") z =" F0167 ' For 77; ( Cir/11.44%”); re WM 0) " D cake/1 x3 oer/£91149 d/m mam/95.5.. Nit/5, A ‘ . 7% ¢ / 7:12 m : ./-£é 5” ’7" M“ W" 7?? Sm >24 (4 (mad). Thus, #044 E3. U) Mlfit .g Cansiwaé Ff é/low Win/1‘ AP— —= Cymbal-4x721, seam}; fir m 44A 7m .' [0V filial/0.131, ( 90.100] 0J5”! 0.129”) 51m éf A; mi wasted/1+ ,roz/m 474,175 77,: Vb Varmé/é‘s Med Ar 77m; proé/zm are 1m! @rreai‘. a 7'39 8.542: ‘| ——-~———————~—————r 8.42 Bloc (assume ,u = 4.5 x 1ij5 lb-s/ftl, 50 = 1.0) flows through an artery in the neck of a giraflie from its heart to its head atia rate of 2.5 x 10“ Wis. Assume the length is;1¢0 ft and the diameter is 0.20 in. If the pressure at the beginning of the artery (outlet of the; heart) is is [DH equivalent to 0.70 ft Hg, determine the pressure at the end of the artery when the headfis (a) 8 ft D =0.20il). above the heart, or (b) 6 ft below the heart. As- sume steady flow. How much of this; pressure difierence is due to elevation elfectsrand how much is due to frictional elfects? (1) _ z. I 1 $5332, -: 4% 175’; +l%{%, Were WVfiV (I) and. 5 5H3 VD =%=%§éz 31=/,/‘5‘6:££ 721/5; Re=-€jr, or I; “if” ‘ . £215 1.44%) “H . Re = = 8 23 flame, Me flow I! Ian/bar wi/b 4.5x! 'fi; -451 .. 54‘ .. 7(— Re "- 82.3 " 0'0778 Iq/so) WI 2:332}; =(86‘7g,)(0.70f1‘)= .593 £7.53 Hence, from [inf/J (Pry, —z)"(2;»z,)— f—fi-atevz 0) MW; 2'1 '2', = 8H, lb 5 _ 0H 3/ 2 ff 593-”: {meme} 0.0m (9M 7350.191?) = 1b. _ lb , A .. > 593 Hz 4995;; 59,531 « sags—#2 A/o/e 5‘4‘7ff4fé is clue fa e/evdian’ — 515 119 due fa 1%‘67‘1'05. b) Wif/i 22.“?/ = *6f7" f2. = 593 7/31 “(523‘ #:X-o’fll) — 0. 0773 (-50,” align“?! gt): #1 b i » 49.31%; 4 379: ,’—,.s. £25345; = 90914?- A/afe 1 379‘ #1 is clue to eleva‘litimJ “ 57-5??- is due fa irriczliafl. 8—3? 8.66 Water flows steadily through the 0.75-in. diameter galvanized iron pipe system shown in Video V81} and Fig. P8.66 at a rate of 0.020 cfs. Your boss suggests that friction losses in the straight pipe sections are negligible compared . to IDsses in the threaded elbows and fittings of the system. 6 m' 'ength Do you agree or disagree with your boss? Support your an- swer with appropriate calculations. 90° threaded elbows 6 In. length 0.60 in. dia. 4’ Reducer Q 1= 0.020 cfs —1 in. length 4 in. length Closed ball valve Major loss =‘ where [:5 (6+6+’7‘+/)in. = /7/‘/7§.J 0: 0,75,;,_ I FIGURE [38.66 .750% R £10.: 525’”? 27-44 6 V mum's?" = {ix/0'3 (see 735/6 8:0 we 067‘th (see Fig. 3.320) .— = 3.37200“ anal ML“ — .113 J I .~_ 0.038 so #m‘ {£125 “ 0'038 035,». 7.; ‘ 0"?“ 2;. U 1 ‘ Z ‘2- ’ i . .1- _, V Minor 10:; =2KL3‘? =[2 + 2 +0.15] 2;: — 5/55; (2) / o ' A2, (3.6/0. 2_, 90 elbow i ["6610ch WI‘I/I 77l— ~0..59‘ (see Fig. 8.26) 7771/5, from £75. (I) andfzjg ma'ol‘ [0.98 _ 0.36] z _ "mi/20)“ 70:: _ - 0./67 = MJ’Z 2-} : Probab/y dining/‘66 (4/175 50:: became Moe {Nb/I'M ll? 060127! /77.. of olher lanes. : 3‘58 8.71. Water at 40 °F flows through the coils of (1::>\Threaded 180° . the heat exchanger as shown in Fig. P8.71 at a »- ' return bend rate of 0.9 gal/ min. Determine the pressure drop » between the inlet and outlet of the horizontal ‘ ‘ device. \ 0.5-in. copper pipe (drawn tubing) ‘ FIGURE P8.7l %*%§+é =4£2¥§ +22- *(f% +5K,_){;2, Mere 2,41, 9 2’ w “PX-1%,? (—11%) Thug {2,702 = (7% +Z/t)-2Lei/‘, M“ I: egg—H) =/2H and 2/65 70.5) =/0.§5 (see Table 8.2) 19/30, from 755/3 8.! % =§(0.ooooos H/(o.5//2 {4}): /.2 X/O-l" 5va _ (1.441%;9 and - [.66‘x/o'5gf "17" = 3670 {see 7215/9 8.! For 1/) Hence, from Fig. 8,20 ,0: 0.041 404/ from 5g. (I) ’0, 701 = (0.09/( gig/fl )+/o,5) ('21:) (1.94%?)(M7ét )2 8-75 T‘T "‘9 'm ' i 3.75; pump shown in Fig. 198.75 datfihfis»ah?$¢ af 2350 ft to the water. Determine the power that the pump” adfié mm was ter. The difference in elevation of the two ponds is 200 ft, Lnlbaw Pipe length = 500 ft Pipe diameter = 0.75 ft L 3 Q3 Pipe roughness = 0 a; =/2.6’ f'” 2f _ or i . Hz ‘ " Re = $6.22 X/éyv ' and from 5'7,-8.20: ». .* ‘ i Re ' .- (1).. . . (2.) _ 7 g 7774/ and ermrxo/m‘im. Aways f: 0.02 - V=.//./f-’-‘ “g9 5;, 7m" a_ JEL,fzomm¢amz g \ j i % E ‘ ' (I V‘tz ‘ ,1.» j ‘i Amme {sac/zei- l/= may “L A’e v7.7x/a‘t-v f: 0,2/2/cia.o/z;g r wVT/Im; V= /2.¢é" ml 3 “Z *5 3% =(‘M ifs)azlvisfifazwfixzwm =assx/fa“f§‘i , -/£ /) -.= assx/v" ill/7K fl = Lari/0. c Alfernah'Ve/yj we could rep/ice; £713) (#73 Mona/y amt/'1‘ ) by 35 ’ , _ ,_ {WW ,. . A » , 8—67 8.75 , (can?) Hhe Cb/ebroak eqUaHan). and? obfafn Va: fol/om. From Eq. m" Vs 32.20/(6d31f +123) y" Mic/7 when candy/heal mil/1 £12) [116: J i 7 7 (4) Re = 6.2zx/o”[3zza/(667FH2.60F :: 3.53X/0’5/(657f1‘12ffi film, #76 Caleb/wk aft/fife” 5/1) :01} .L 2.5/) (s) W = —2.0 log W By combining Eqs (Hand {5) we Mia/'17 a s/fly/a eymh'm lbw/me 012/}: f: .L g _ A2.sy(667yf+12.8)%] W 20 [0917333 Xl‘bé I Using a oompufe roof' fiM/ng program ‘lo salve E7 (6) 7/1/03 10¢ 0.0/23, cam/Hen! buff/7 #76 above fn’alanal ermr muffled. 8-68 ‘ 8.92 A fan is} to produce a constant air speed of 40 W3 throughout the pipe loop shown in Fig. P892. The 3-m—diam- leter pipes are smooth, and each of the four 90-degree elbows has a loss coefficient of 0.30. Determine the power that the fan adds to the air. fl FIGURE P8.92 L _ We [0647’s (/Mlld (2-) 0/ ME mange {lo/ace {aflom f/yaf Wsz 1444/1/76! 2/42. 3 hoe/t =(f%+§:l<ti>1‘§ Mere IQ. = macaw, :fl/soJE Re :— E—KQ = /.23 %{%¢a%){3m) : ‘ [A /.79x/075_,;£ :ahalff—so :0 Mm? m 59.00.20, haem- .—.— aux/o“ Hence) ( Alf v __ f (20+2.o+/o+/o)m.- #05 l/ m 1/}: — (0.0083 ——————3 m ‘ M2) Zane/Eat) / * V so M * ' w arth -—- paw/g, ={fl23'g23J’{£o°/%)[ $r3m91rszvg—fl mm: 3.77x/054éa = 3,7421% 8"88 ...
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