me320-hwk5-sol - The University of Michigan Department of...

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Unformatted text preview: The University of Michigan Department of Mechanical Engineering ME 320 - Sections 1 & 2 Homework #5 — Due October 19, 2011 Problem 1: Problems 3.46, 3.88 and 3.90 of the text. Assume quasi—steady flow in all problems. Problem 2: Consider the reservoir system shown in the figure. The reservoir is initially filled to a height H with water of density ,0. At time t : O, a valve is Opened and the water begins to drain through the disks at the bottom. Neglecting viscous effects and assuming a large reservoir (so that the velocities in the reservoir are negligible), obtain a differential equation governing the flow rate Q(t) for t Z 0 in this system. Problem 3: A sanding operation injects 105 particles / sec into the air in a room as shown in the figure. The amount of dust in the room is kept at a constant level by a ventilating fan that draws clean air into the room at section (1) and expels dusty air at section Consider a control volume whose surface is the interior surface of the room (excluding the sander) and a system consisting of the material within the control volume at time t : O... (a) If N is the number of particles, discuss the physical meaning and evaluate the terms DNSyS/Dt and 8ch/8t. (b) Use the Reynolds transport theorem to determine the concentration of particles (particles / m3) in the exhaust air for steady state conditions. (c) Consider this problem for the case that the ventilating fan fails to work so that V1 : V2 : 0. Evaluate and discuss the meaning of the terms DNSyS/Dt and 8NOU/8t for this case. ‘ ‘ F ' ' ' * I v 1 * ' - ' v * 1 + - I ' I r - r I v r I - - r - - m - a a - - I - - - - . + . I 1 I - r I r I o - r a + . I . . I I 4 a . . . . . . . r . . . . . . . . . . . . . . . . . . . 1 . . l . , . . . . , . . . . . . . . . r - - - - - - I i s v a l r . . . t - u I a . . . . I . . I . , . . . . . . . . . . i . . _ _ . _ ' _ _ _ _ ' _ _ _ _ _ _ *I-l:i:lllll:HJ1-I'1:r:I‘I‘J‘l‘i‘l.lIi'l 15-min-|I.|Ia“Ir-II.I.91‘.I-:..‘:|-hl:1-:I:-u-:I:+::.:;:.-:I:n.:.:.:.:.I:'..-_:.:.Ir»:.|I..:.-:.:.'lI-I.Iar...I..i...II.u.'_.I.:..m.I“.,+,,.:.:..II:‘I;.':filial-F:If?”:II_:.‘II_:.II"-:.I:_:I:.I'I‘:,-:I':_:.I_:I-Irlll“I'm”...-1 3'1"..." "iii. _ / .',4_ F I - I i - | I - - ' I 4 I I I I 3 I I - I-- - r - - - 4 p h - I a - - - u - - - - 1 n - p p u - - . p u - . . a . . . . a n a u . . a . . . a . . a . . . . . . . . . . . . . . . . . . . . . _ , . . . . . . . ..._._. l 2 I I I E; n-"r' ' tr. ln... .' .'. .:.: .I- . .I_.I . a in.) I I I...- 1:: o‘- .. my 2 fl? A2=O.1 m “M L. Problem 4: Problems 5.13 and 5.16 of the text. Problem 5: Problems 5.26 and 5.31 of the text. 3.1M Pop (with the same properties as water) flows from a 4~in. diameter pop container that contains three holes as shown in Fig. P336 (see Video 33?). The diameter of each fluid stream is 0.15 in., and the distance between holes is 2 in. If viscous effects are negligible and quasi-steady conditions are assumed, determine the time at which the pop stops draining from the tap hole. Assume the pep surface is 2 in. above the top hole when t = 0. Compare your results with the time you measure from the video. Q=Q,+Qz+03=h/qrg{l aFiGuaE P345 were 0‘1 M- = W- a- and x», a a anew (£51,213) = /.227X/0—y7C/2 —_'Z.7 i 1.. 2 Th”: HT v 4 {,2 fl) .. o. 0873 {J J +10; 4%] .747" 36%.; when: htsth2=h+LJh3:h+2/' and L =2 I27. HBHCGJ —(V¥A,/Ar)jfn =f0 A II dh . . . - wfiere I! I: 71/79 Ill/979 7’ (W7 +Vh+z WW) {an fez/"1% {mm/trace 1‘0 recto/7 7% V/a/arfia/e or L (lye-a ’ f = A7 a“? 1 Mr 0(V/7+Vh_+2+ mu = 0. as 73 H’- L d1; (/.227XIO+H2)[(1){32'2“3:81;ng +1417 W14 +2.1. ) 771m; 1. of}; 2 )1: b ‘F-‘H50J667 a (VFW/97PM) W m ’2 {’1 Mlle: lat/x76 if» 7426’] 1%}; email/'0» 9/1/63 15 I}? sew/talc. (600’!) 7720 numerical valve of {he 1711'69m/ is ebb/fled [11/03/779 7/90 {rapezoidal rule since #70 closed form ana/yficaf 30/1175an is 17011 given in 1771’199/‘4/ fab/es. 7Z6 EXCEL sprang/35.99% wed {or #1711 is 911/00 lye/ow. L I 7 = 01.7 5 10140 were 101= O 4-. 00.7%; (1‘,- +£1,)(h.-.,-—h.-)] =(6’8-7 fiflo-IwVfik I075 n, in. h. 11 KM. 11ft"2 (172)141-0010“ - h1). fl” 7 0.0 0.0000 1.015 0.00004 1 0.1 0.0003 0.914 0.00743 2 0.2 0.0107 0.070 0.00711 3 0.3 0.0250 0.037 0.00000 4 0.4 0.0333 0.010 0.00005 5 0.5 0.0417 0.700 0.00040 5 0.0 0.0500 0.704 0.00029 7 0.7 0.0503 0.745 0.00014 0 0.0 0.0007 0.720 0.00000 9 0.9 0.0750 0.712 0.00507 10 1.0 0.0033 0.097 0.00575 11 1.1 0.0917 0.004 0.00504 12 1.2 0.1000 0.071 0.00554 13 1.3 0.1003 0.059 0.00544 14 1.4 0.1107 0.047 0.00535 15 1.5 0.1250 0.037 0.00520 10 1.0 0.1333 0.027 0.00510 17 1.7 0.1417 0.017 0.00510 10 1.0 0.1500 0.000 0.00503 19 1.9 0.1503 0.599 0.00490 20 2.0 0.1007 0.591 21 Sum of column = integral = 0.12011 Thus. t = 88.7*0.12011 =10] S 3-92 3.83 A long water trough of triangular cross section is formed from two planks as is shown in Fig. P338 A gap of 0.1 in. remains at the junc- tion of the two planks. If the water depth initially was 2 ft, how long a time does it take for the water depth to reduce to 1 ft.? FIGURE P388 :—-%+V2-2 2. fi-lu-VL'FZ '5’?“ +32 (1) (I) J” 2; ’ :0 J 2, =5, and Ez=0 h fl/so =lé/42 01‘ since I>>flr [7‘ d fallow“; follows fiai‘ 1/, << 14; J Wéere Z = " a? 777st £310} git/85 lé Lil/2; So #7677! _ '1g.§l=,qz 23/, Wff/g 4=5Z=266 dfld 142:5:4/ w/aere b is Me fame? flangf/i- or_I {h a“! = — HIng bub/d? can be fnfegmfed 1‘0 9/1/19 hf] 3’96 3.90 When the drain plug is pulled, water flows from a hole in the bottom of a large, open cylindrical tank. Show that if viscous ef— fects are negligible and if the flow is assumed to be quasisteady, then it takes 3.41 times longer to empty the entire tank than it does to empty the first half of the tank. Explain why this is so. a=av =a‘afl-v =4“ (—3?) where Infeyrmle from 5“” at {=0 7‘0 /2 ate: hath d1; HEW "We—Ma” or ’1 1 21W "@651! H 0f 2 2 a We) [WT-WT] 77211.; flee/Mm? the fan/g /= - (3W1 [3:0 and fa lmlf empty 2%? hulk) 26/ =43” 2. (0): free ratio“: uh. +1? (warn-'7‘ (i)! 384%) Jim: , rifLJ’-GJ‘ 911-8. Brio! I a} W‘IVE' (04 r: r.) A4" mil SECJfOA. (5.4 rzg) 'u‘ we ‘9 fl 2 ' z 5‘ 2 zmxwt .. £+A¢3y+53145 3" 2 g 2 3+ EAL R, E m awrflen‘ao V950 m H» NJQIVNV. - '3. 3(2a“%;)7— iii—.9 g E‘naf-zH(+} o l 'é‘rko tév “La 1 3.! a! ‘3 7:0” <f- __ Jr .. __ §3V E“ i ' ([34 ' SM Jr ,3?» J" {L‘fi'J-f- MIWVoJV s q / Xx . '3 X ,__ as QR \ M Maw + Mm Mao ;/ a . I /’ c s 5 I K I I3 C V \ \ v. N {I} \fl’fi\v/‘u._,~,,flwn ) \ a 'j\ Ex?» I a C \xx» ' ’(LOEF W 4 ' VI . a ‘2’ ( VI ‘\\\—v'/W‘_,r’ Chats?! L M) . (30% m, c vim 7» v1: Jim» W; mg): 'L $31] "5 “ L z \X/Leffi 3—K 7 L fig , 3* ' anLI /: i9; (55,, \ “Q'Rmumg 4+ 2,214 2 r M22223 /. (fl. 2 i9 + ,9 (“Wag H0») ) HIM/x}!sz 3:” 5m“ ZQL‘ 0H“ ERZK'LL’Z 0 +20 Pleem 5, _ A sanding operation into the air in a room as shown in Fig. jamounx of damn; the mom 'is-mainwimd gt a" can? by: {mm draws dammit into mermam In (I) and expel: dusty air a! section (2). Consider a contra! veiume whose surface is 111: interior surface aid-la room («angling the mast) and a system consisting of the material within the come! volume at time I In. 0. (a) If N is'lhe number of particles. discuss 1h: physical meaning of and avaium the W M'me and an/at. (1:) Us: the Reynolds transport them-cm to determine the concentration of panic!” (pani- cles/m’) in the alums: air for Steady state caudifiqns._ mm: 5 >: . m‘ pm‘liu’trh 'mJ.‘ nu... _ 2 .DM 5 d)’ D =-' him rafe of change cf Ha: number of purities I}? 1%: V2-2mls a Azio‘l WI2 .sysfem. flf flare #0 Hie sysfem 5mm: of Npm‘icles. In fact H73 sysfem is +53“ paffgbles for all Hm ism. ' fissummy Hid {be par-{ides do 1101‘ gef "gleeffaydbor, IV remam: (:0me- Thus ,V 31%“ *0 - ' ‘3'?th x gm “"9 0*. 5119099 of H78 maiber of parh‘c/es 131' 1%- confra/ pro/ma; Depend/395: an 1%: Mia a-f win“ {he Sandw- crearespari‘ic/as and ejem‘: Men Ma Ma mm compared 1‘0 fhe raft: m‘ whiz/I {be fan draw: flam {rm 1%: mm, we could have 95%! % 0. . N I 5) gal/jig = 35-53 +179} rd: of Hoard parh'cles am‘ of Mm} volume or {or fiend); sfale éfifiw Sg‘Maf: Haw mt parficles info awry] valvme (f/m sander, none enfer I41) - Flow of farm/e: am‘ of mnfral viz/am ( filmy/a I’m aha/:1" :91 7720.3, “’05 fiffc/es_ 3 I flam- J- Mm»: ni= parfiqle iconrzmlrafion (matski) Hence, rib, ‘ . . muff-“J = 5m; gar-£15k: // CC) DMZ; : O L ( 9mg we wumbgy of particle; Yeww‘ws (ow/472;? ’ 9* prov‘de paflft‘cle; d0 WC bewwe #5516164" @Wev . 0*, 1%va “We COMJCVQI Volume WW4 \A =szo 7 We only How 0(CV099 we cox/Jim! garage (‘9 firm We gander 61W! F9 eapmf “to ,, [05’ lam/tides '9 - Thu; 7 856/: U) become; I BNGV. _. 9 E’fffides) (BNGV \ 5 fl 0 __ . 0 Y ————4\ YlCEf 2+ H I 9 ' 0 M “9 fl;- Wet air m = 156,900 Ibmihr 53.13 An evaporative cooling tower (see Fig. P513) is used to cool water from 110 to 80°F. Water enters the tower at a. rate ’of is 156,900 lbm/hr. determine the rate of water evaporation in lbm /hr and the rate of cooled water flow in lbm/hr. ' Dryajr —)- rh =151.0001er For sfeady flow of 00y air “F‘IQURE 95.13 If] :m (I) 3 2.) d1? airr' Far H2009 F/0w of Wafer ‘z' (I- l 2’ wafer ’1/50 = ' + ’1'? m2 m2, dry 0,)- 2, wafer- (3) Coméimhj E135. l and 3 we yef ' = m « bl? = I’d/’6 0/ wafer euaporab‘m—r mzzwaier 2 3 50 s 156 900 “’m _ 15/2000 L4: 0 5900 L5: -.-._— m30W4‘lo" [7,. hr Ar 50m £3. 2 we 96+ - M —— rid = I‘m‘e 0F coo/ed wafer flow? 4 H 1 2,w0+er 10,, M», __ 5700 [£0 ._= 244/, 000 __ g: [Ir hr“ 5-!6 5.]6 Freshwater flows steadily into an Open SS-gal drum initially filled with seawater. The freshwater mixes thoroughly with the seawater and the mixture overflows out of the drum. If the freshwater fiowrate is 10 gal/min, estimate the time in seconds required to decrease the differ- ence between the density of the mixture and the density of fresh water by 50%. A {raced , nan- deformriy confim/ Vo/uma 741ml cmfmhs 1%: (h the 55 - 94/ drum 1': used. Fresh wafer confra/ I’d/um: wafer- mixfure err-hrs MC wifh densffy, ’6’? J and volume flowml'e’ ’ - The Ml'x'f‘urc i5 dimmed 7’» 15¢ homageneous +hroughou+ file. General Volume and leave: Me Camfi/ Vd/umc wiH: density, 3 I and volume flown/rach az. App/,‘cafi'an of five. Cmatcrvafian 9/ mass eqaafi'an 5-5) 2‘0 7‘68. f/ow f/trauy/q 1%"; Camt‘m/ Ila/um: yields _ = {/J 346” t 4% M 0 v 7,“: ,. are, IhCOMp/GSS/é/C’ Q: = Q; :. Q . WM} Volume. is msv‘ztnf- 771115 slhce 'H’IC. {/m'ds' (Java/wed Mia, fhe. volume of M £35.! lead: 1'29 t: 1" g Q = Q di . or 4’?) + (L) g ., 2-1 (2) CH ,4 47ch {73/ 7776 Jolufian of Q53 2 i5 6: = C 5‘wa + 10 (3) /.9 ' 5/117: M f: 0) = freame , {’99 4'3): [026 I ffiesémhr (/ 94 mfg/#3) Hour C = 0.02é (can't) " 'fm'a/ ml‘KfiH’e densi‘fy im'w‘ia/ mifiwe density fubsz‘h‘uf/n'j flu‘: m/ue (fggl/ngcrff/Vm‘ "3"? 53-3 W3 3” (55y!)(6ar§§ag) - [.013 = 0,026 e + [-0 and 1E :— 22? s 5172 5.26 Estimate the time required to fill with water a cone— shaped container (see Fig. P526) 5 ft high and 5 ft across at the top if the filling rate is 20 gal/min. I FIG U RE P526 513m Willa/7'0!) af‘ #4: (mien/47245;: 6% pics: pr/hcxjo/e 74? {£5 can 'fml volume. Slum/n 1h flit £39qu W6 lam/a +f/afi3dfi = 0 at cs For I}? compressive flow éi“_a =0 9t 0r 1‘ f [w = Q/dt‘ 0 0 Th 2 1': "5 Jo; 72m A , 7f (gawk; f+)('728 a") 5 f g a 1261 ’ (121(20w)(23/§_’-3 ' 3 Min ‘7‘! f = [2- 2. Min 5-3! Storm sewer backup causes your basement to flood at 60%”, I Volume find- the steady rate of 1 in. of depth per hour. The basement floor f canffifns Wafer area is 1500 ftz. What capacity (gal/min) pump would you rent —-— —— -- — — — — — — I I0 (a) keep the water accumulated in your basement at a constant //I x I : h level until the storm sewer is blocked off, (b) reduce the water / i / / ] accumulation in your basement at a rate of 3 in./hr even while / - . _ / ’4 “ ‘1' the backup problem ex1sts? - [‘br 57 661610??an Confi/o/ Vii/«me flmz‘ can/67x37: J‘Ac wa/or over fhe Ansemenf floor- (566’. 51$;th agave) J fee Conserva/a‘m 09‘ mas: pram/we, (.53 52/7) Ami; f0 *ffll-f-fia/fl :0 at CV C; or 761» Cam’fanf f/m'd dens/7 and area (A) d-é “ (4) 5r Pan‘ 6?. J / 4344576)- Qflue = Q5? 729 eva/uml: 4;,“ we use, 53'. I wi/t, QM-so. 771m, _ = A (111 z. {1500 f¥‘){/ 31. / at»; df hr)//Z )= 125 19‘? . 1‘1. 1” and fl = (/25 fig/29485627 _/_ __ ,56 34/ Q0101" ;,,. {:3 “fl " _'_ I”; In- (b) For pow-f" 6) 53/ yie/d: Q0114 = Q"; F A Q o’t =15.6!3_’ _/ 0029‘ -3/_11- (Tfififl)/J__ Q0“. mi” X hr)125.) {1,3 60,??? Q + = 62.4 22! 0“ _— Md}, 5-25 ...
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This note was uploaded on 11/22/2011 for the course ME ME320 taught by Professor Akahvan during the Fall '11 term at University of Michigan.

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me320-hwk5-sol - The University of Michigan Department of...

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