PS6_Solutions

PS6_Solutions - 531 Water enters a conical diffusing...

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Unformatted text preview: 531 Water enters a conical diffusing passage (see Fig. P53) with an average velocity of 10 ft/s. If the entrance cross section area is 1 ftz, how large should the diffuser exit area be to reduce the average velocity level to 1 ft/s‘? Section (2) lOft/s V2 3 lft/s 1&2 FIGURE 95.; H‘C ll ll 5r Shem/y m candprfisflbk: "LT/aw éci‘wcgn gecko,” (gandflz) 5.2 Various types of attachments can be used with the shop vac shown in Video vsz. Two such attachments are shown in Fig. P5.2-—-a nozzle and a brush. The flowrate is 1 ft3/s. (a) Determine the average velocity through the nozzle entrance, V". (b) Assume the air enters the brush attachment in a radial di- rection all around the brush with a velocity profile that varies linearly from O to Vb along the length of the bristles as shown in the figure. Determine the value of Vb. ' 77m) AIM:Q2 0" 14:”sz so 4-7 V” = 45.9% (w%=ntmmnpvélm4@:2@wmm lg = aye/"age Veloc/i’y ml (3) = 3L” Vb and ,43 = 7703 63 T/IVSJ 3 swwsntem =/ or m=2a4§ ‘ 5.3 The pump shown in Fig. P53 produces El steady flow of 10 gal/s through the nozzle. Determine the nozzie exit diame— ter, 1);, if the exit velocity is to be: V; = 160 ft/s. IFIGURE P553 - w I in? m3 a For sfeady How QI—Qzl where Q, :Iaigq-(zmmflm) = 1.337% Thus, wfih vz=m§fi *3 1.337%— zflzvz =%D:(zoo%*-) 01‘ D,_ = o./3ofi= 2.57 in. 5.4 Air flows steadily between two cross sections in a long. straight section of 0.1-m inside—diameter pipe. The static tem— Section (1) sects‘cm (2) perature and pressure at each section are indicated in Fig. P54. If the average air velocity at section (I) is 205 m/s, determine P1: 77 “Pa (3‘35) p2 = 45 “:3 (abs) . . . T = 268 K T = 240 K the average at veloc1ty at section (2). v1: 205 m/s 2 I FSGURE P5.4 7/7}: anaéw/x i: JIM/Var 75 #1: one a?“ éi’dmp/e 5’. 2 . For sfead’y f/ow Ae/Weeh 'Jccr‘l'ans (I) ana’{2] m . n S .5: l - lee. Asfum/nj uflm; under flie Mimi; 19m; 0/ f4}; fmé/em/ m}- Aehmxe; a s an idea/ 945 we: use #16 z‘dea/ ya; cguafi'on 0f sfa/e #3) 79 967‘ = (2) “ ’2 ’3 7, (Oman/Jay €55. / man 2. and abserw/la; Ma)! 14, =41 We. gef Vi = P; 72— V, = [-feflms](240/<) “<5 H 5.5 The wind blows through a 7 ft X 10ft garage door opening with a Speed of 5 ft/s as shown in Fig. P55. Deter- mine'the average speed, V, of the air through the two 3 ft X 4 ft openings in the windows. 16ft I FIGURE P5.5 For Sfead‘y 0‘7 compress/He f/atv .— anmge '- Qw/édaw + @00de ddor 0* A V g 7-1 V + 4 V 7mm Mm; +v Mr.de W/nde dddr fanatic, door so fie am}; glgeed') V) m‘ fhe d/r Mmujk fire *wo [5 V # Agme mm; +1, Wm w _ C714) (70 760(5 4?) mm = 4,9? ZANlr-IIIOW I, L 5.6 A hydroelectric turbine passes 5“ million gal/ min through its blades. If the average velocity of the flow in the circular cross—section conduit leading to the turbine is not to exceed 30 ft/s, determine the minimum allowable diameter of the cOnduit. For xhcompresfié/e f/ow flimujh Me candor/f and furé/hc .— - lander/'1'“ fare/me 73a; A V .= Q . candalf candy/f furbme and ._ ' d = :4 EMMMC == Corlde = m H dwmfiu’f i— 5-5 5.7 Water flows along the centerline of a SO—mm-diameter pipe with an average velocity of 10 m/s and out radially between two large circular disks as shown in Fig. P57. The disks are paraliel and Spaced 10 mm apart. Determine the average velocity of the water at a radius of 300 mm in the space between the disks. For sfeddy I}? compress/6!: flew ' (4)277 '71 _ mzdwoommom) 0” Q, g 82 ffi=10mls AV, z 4sz FIGURE P5.7 Thus 1 ._.. ‘2. \7=A.V,- 779- V, _ (50mm)(10m/s? z __ dz v1 = 1.04 g: 5.8 A hydraulic jump (see Video V105) is in place down" ,j stream from a spill-way as indicated in Fig. P5.8. Upstream of the jump», the depth of the stream is 0.6 ft and the average stream velocity is 18 ft/s. Just downstream of the jump, the average stream velocity is 3.4 ft/s. Calculate the depth of the stream, 11, i just downstream of the jump. ft/s fiflsm'sj ” " ,1. ' ' 7 - m, I FIGURE £35.60) 0r_ fl VA, = ngAz Thus W, = V2111 and A2 Viv/4. = (/5300261‘7‘) _ 31mg 1 gTT—m— /‘"wmm\ 60/1790, va [u me, 5.? A water jet pump (see Fig. P53) in- i volves a jet cross section area of 0.01 ml, and a jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross section area as- sociated with the jet and entrained streams is i 0.075 m2. These two fluid streams leave the pump ‘ thoroughly mixed with an average velocity of 6 ‘ . .m/sthrough a cross section area of 0.075 mg. Determine the pumping rate (i.e., the entrained I fluid flowrate) involved in liters/s. Far steady in camp/essi/a/e 790w firm/(9h 7%: com‘w/ Vo/ume a + a = Q3 a2 = Iva/43 ~ W, = [(6%)(0-075 W“) - (30%>(00'm3]tm$§ 5- 5? 5,10 530 An evaporative coating tower (see Fig. 135.10) is used to cool water from 110 to 80°F. Water enters the tower at a rate of f 250,000 lbm / hr. Dry air (no water vapor) flows into the tower at a rate of 151,000 lbm/hr. If the rate of wet air flow out of the tower is 156,9001bm/hr, determine the rate of water evapora- tion in lbw/hr and the rate of envied water flow in Ham/hr. _ We: air m = 156.900 lbm/hr gecfian (2 ) Warm wars! —> .. ti} = 250000 lbmthr Dry air —> rh = 151.000 tbmlhr For Sfeaa’y f/ow of diy air FIGURE P5.10 M3 : m2) 0'43; ah” (I ) For Sfeady How of Wafer m/ : m2, wafer + 4 (Z) I4/50 m2 = m2, dry air + m2, Mid/Er (3 Comb/Why E35. I and 3 we ger‘ n}; = m —— n33 ~== [Mfg of wafer empamf/mw 2/ MW ,2 50 I I6 - s /5 00 Lin -. 15/ 000 [£2 =— 5900 .5: mmwdlcr 6’ 9 5,, ’ hr Ar Hom 53. Z we 96+ m = Ii? ~ 105 -= rmLe of coo/ed wafer flay q / z,wa7‘er or [6 ' g 250 000 IL» _. 59700 1%: _.= 24165 000 .32 m 1/ ’ ,5, hr hr s—-q 5.11’. At cruise conditions, air flows into a jet engine at a steady rate of 65 lbm/s. Fuel enters the engine at a steady rate of 0.60 lbm/s. The average velocity of the exhaust gases is 1500 ft/s ' relative to the engine. If the engine exhaust ef—- fective cross section area is 3.5 ftz, estimate the density of the exhaust gases in 1bm/ft3. 5r steady {low I773 = I77, '7‘ /7’l‘Z 0F 5—10 secfibn 6') (/h/ef ) 5.1 '2 Air at standard atmospheric conditions is drawn into a compressor at the steady rate of 30 :m3/min, The compressor pressure ratio, pm/ pink“ is 10 to 1. Through the compressor p/p" ' remains constant with n = 1.4. If the average velocity in the compressor discharge pipe is not to exceed 30 m/s, calculate the minimum dis- charge pipe diameter required. 5.13 Two rivers merge to form a larger river as shown in Fig, P513. At a location downstream from the junction (before the two streams completely merge), the nonuniform velocity profile is as shown» Determine the value of V. 0156 fire. Cm’fl’YD/ VOA/MC 5fiywn wfifiqn'n brake» //’)’)CJ' I}; 71106 Ska/$51 above,- We maize flaf Mcloflv and Mn W CWSCNfl/fi’an 0/ may: IUYI‘nc/‘p/e WC 9‘67“ ' l 2‘ '3 0.3V v 7M: my, mall/2 == {0/2832 v + [7 Av \/ and _ . \/ =- 42”; + ’42 l/z. _(50H)(3fl)(31g +{gfi’fi-{04fl ,4 (m + ,4 (yoffiéfljflmgfl (705%2‘7‘) 0.9V V V s 963751 .5 —_ —————- 5"42 5.1% I ' 5.14 Oil having a specific gravity of 0.9 is pumped as Section (1) illustrated in Fig. P5.i4 with a water jet pump (see Video '\"3.6). The water volume flowrate is 2 m3/s. The water and oil mix- ture has an average specific gravity of 0.95. Calculate the rate. Section (3) in m3/s, at which the pump moves oil. _>_ water Water Q} = an? « "Lm3/s “‘3'” Section (2) (8G = 0'95) Oil (80 = 0,9) For “eddy {my FIGURE P514 . . z ml + fl’l2~ 3 Of“ flqflk/fQLz/gqa (I) 14/50 Slhce, #12 Waller and 0// may be. cow/'45er ,,‘, Com/Wx/é/fl / Q, + €22 = a} (2) COMé/n/nj 5?; /ama’z we. gei‘ pIQ/ + 542 = gamma) 0f” , Q; + SGZQz = 563(6)! + Q2) and Q n a, (1— 563) z _ 5G3 - SGZ Thus 3 \ Q = (1%)(/”01%_/ .—,- 2001": 0.625 r 0.70 5 O 5. 15‘ l WW“— Air at standard conditions enters the compressor shown in Fig. P5.15 at a rate of 10' ft3/s. It leaves the tank through a 1.2—in.—diameter pipe with a density of 0.0035 slugs/ft3 and a uniform speed of 700 ft/s. (a) Determine the rate (slugs/s) at which the mass of air in the tank is increasing or decreasing. (b) Determine the average time rate of change of air density within the tank. m 2:»: r U56 the va/umc Wif1”I;’), M6 low/Zen Air/165. (a) Fray/s ‘Hm’, (macaw/4790M all WSI pnhclk/t We wy‘:M.-—Vh asia’aiaofl V pf . m amL I}: In “7' 0“" “ff 1 D MWI — @0023? f/uj (a f+j_ @1925 353.9 77 ["2 "0/700 353‘) W ‘ ‘"‘ m 1H 44 if pf- .‘ff'3 5 ‘ / ff>) 24457; = 0.00156 {9f 18' crew/£37 Dz‘ "' 5 (b) 9.44:7: WP '57: D: ‘4” DE _ 0. 00456 5;"! at 5,:- Dz‘ / J} 50 D 0.00 ‘HK If,” __ 0.00954 {try ._. 2.29xm fig 4 z: '- 3 pg— V 2.0 7"?" I}: O p 5" ” 5.16 An appropriate turbulent pipe flow velocity profile is R... r‘ l/nA V: i ””( R where u, =Acente1'line velocity, r = local radius, R = pipe radius, and i = unit vector along pipe cen‘terline. Determine the ratio of average velocity. E, to centerline velocity, uc, for (an = 4, (b)n I 6, 02):: = 8, ((1)12: = 10. cross Sec fizm area For any Cross sad/On area. h'v=/0Au = {/2 Mme/A fl/so .1 _.3 ’1 we R~F . z o = M .__....... Vn \/c c ) 73a: for a um'zérm 4: cr’llrz‘r/éufed densify/ fl Over area A I? [K n 4 ac (Rf) Zvrrdr Z TI: 77"? 5.17 Water flows steadily thruugh the control volume shown in Fig. P5.17. The volumetric flowrate acmas section (3) is 2 fig/s and the mass flowrate across section (2) is 3 slugs/s. IFIGURE P517 (a) f 6‘ VI? 51/! ==~weigl1f flawrafe across area (I) Lyn}, m (/Vofe 5 ‘5‘ am < 0 since ‘7'»? <0 for Mei/917W area (1)) m B y conserv‘afion of maSsJ for arfeacly How) "a, r- ”;2. +Ih3 1: "52 + (33 Q3 = 35/1/73/3 S/VyJ/fijZ 0!" 0"), =3 6385/0951; Thus; from 137.02, [3' (M :(“32.2 H/s‘)(6.ee Slugs/s) 5‘ " ZZZ/dug ' H/s")/S 3’ :222 MA: (I) (b) I V9 V513 M = Momenfdm f/w: amss area (I) (I) L ' 0/7 (01 fi=+c:053005'+sm30f and V,‘ \{("603300? ’Jsifl30?) =‘Wl31, where 9/], 5/75, r-6-W 35% (from 5712)) Thus , V: m, v 6.33% __ ’ 6’7" (I-9*if%?)(a.mf) Hence, WWW/H Wemfi.)~h‘,/I, =-€\4M=-»'v,14 “may (I) = (6.88 %£)(8.87§)(59330?4 91130;“) .2 (52.82+ 30.51) = 8.8795 : 52.8;‘+3o.5; lb .5 “I6 (I) O 5. m ]——"—’”———__' 7N As shown in Fig. P5119, at the entrance to a 3—f‘t—wide channel the velocity distribution is uniform with a velocity V. Further downstream the velocity profile is given by u 2 4y — 2y2, where u is in ft/s and y is in ft. Determine the value of V. we f’he wm’n/ Vo/ume Mal/cake! éy m6 [am/<04 Am"; pa flit Skefolq above- BM #11; cdm‘erwné'on 0/ max! FYI/7690A? A2 Jar 2 “If H3 — 3’ 47,2 7b: 9% __ vMe-ng Z? g}: o ? j 5. 2 o _| 5.20 Flow of a viscous fluid over a flat plate seam" (2) surface results in the development of a region of Section (ll \ U Outer d e ge reducedvelocity adjacent to the wetted surface \ as depicted in Fig. P520. This region of reduced flow is called a boundary layer. At the leading boundary layer edge of the plate, the velocity profile may be r___,,f considered uniformly distributed with a value U. ‘ »-;,, ,,,,,,”,,,,,,,, ,1” I,” III/III“ All along the outer edge of the boundary layer, ‘ ' ' ‘ , x the fluid velocity component parallel to the plate - surface is also U. If the x direction velocity profile FIGURE P520 at section (2) is 1/7 E ___; X U (5) develop an expression for the volume flowrate through the edge of the boundary layer from the leading edge to a location downstream at x where a the boundary layer thickness is (5. From #13 comm/a 1%“ of mass pr/hc/‘p/c app/fed 7b 7%: {/0W 7%,???” 7V“: canfiv/ Volume Shoot/n I): Me f/jm’é we have. ' s m = \7’. "a’A 2 1‘5 r Iiawmpressib/el “flaw I =flVIy g: 7 d /0 QM 0(5) 6%) where I :: Laid/fl 0/ 726 p/al‘é and #1445 = JUN 634/ 2? 5.22 Estimate the time required to fill with water a cone- shaped container (see Fig. P522) 5 ft high and 5 ft across at the top if the filling rate is 20 gal/min. IF‘IGUHE P532 50m aff/iczfi'an 02‘ flc (anscryafibn 0/ was: pr/‘nc/jo/e 25 fit carav‘ml valume, :Aown M Ma {ya/c we have _3 7t. flVqu/l =0 31‘ 0/ CS éi‘ - a s 0 at 0r 1* + [M = 42/414 0 0 Th 2 l 2:3 a: g V g WDA g 7(5;+)C5{+)(/I728 “2) Q. ’2 Q (/2) (20 537/ 23/ [if and mm>( if] 1f 5 mil?) 5.24 I 5.2- L!“ Storm sewer backup causes your basement to flood at defymm‘a 7% 7L ' the steady rate of l in. {of depth per hour. The basement floor f 6500;] Vie/“9‘”: a area is 1500 ftz. What capacity (gal/min) pump would you rem —— -— —— w — — ~ - - a, r to (a) keep the water accumulated in your basement at a constant / A ~ / I level unlnl the storm sewer is blocked off, (b) reduce the water / l / I ll accumulation in your basement at a rate of 3 in. / hr even while / l l l the backup problem exists? /. _ 7.1.“: _: : : t. :l" ‘ ‘7 V , _ , , l / / n n :/ a” ‘1 :éW/%wmm '.-I _ _ J7. __ .I- .. :‘l/ _ {low ml flaw in E?!” a flefirmmj Con/ra/ Via/«me 7‘th Came/37.; 7%6 Wfl/er (mar 7%8 éasemcnf f/aor (see “614% awe} / Me Conserua/fon 09‘ Mas; Mme/we, (53 5.17) /66245 +0 f" :0 at CV cs P or 7%)» Conf/‘an/ {I’M/d dens/)7 and area, (A) A 513 - 62,." + QM ==0 I m (a) '5)’ par‘l‘ 6!, J / brads-7‘0. Quad = Q5? 779 evalualc aim we use, {5%, WW, Q’W=0. .flws . s A W» ; (/500 f+"){/ m. / - 2 ‘2’ "' h7)/ ‘ )= /25 H ” dt /2 31. h: “"4 3 Ff ' ; 125 fi)/Zygyal) _/_ W 56 4/ Q” ( hr ,4? w ’ ___i'—.— it: hr (b) Far parf AJ 53/ yie/d: Q = Q,~ — A «.111 au-l‘ ’1 dt .2 [5,6 in! __ 00,9“ (.3151. (796731 J, QM 55? fig ll hr)lzf') #3 6031.5 f r Qawf = 6——-———2"4 7""! ml}, 5.25 Two 8-ft-wirde rectangular crates are wheeled into the k 6 “:1 trailer portion of a semi-truck a; a speed of V = 2 flt/s as shown ii DA; in Fig. 1F5.25.Attime 23 == 0 the final of the film cram is 2 ft from the open back of thc trailer. Plot a graph of the air flowrate across the open and of the trailer as a function of time for 0 5. t s 10 s. Consider ‘Hve deform/M emira/ volume shown in {he skeia/I. From 5715/7) 3%]941/ '1‘“) W’gdfl-‘fl CV CS or sinus Pficamhm‘ fide+fWIfimzo or $1,? {mafia/xi: -v,// “cam/m Q, = airflow rafe across and of frank/t 19/50, 1 £5 = r‘mlg ml which #73 w/me mpg/1‘ in 7%? I‘M/Var Manges MUM 79m =-V/1m;e J where Harm = cram—sem‘iml area of fl): amfc Md [.5 crass/177 file and of 7%6' I‘M/7W. 777W, — Vflcrdfc 5 ~ 01' 0" Q, 3‘ Vflcmfe J W/Iel'e V‘ 2 6%“ ‘ (I) From 1%: 911/3” Jain z flora}: : 0 7%" 0$£<lsj flank ‘1 {3H)(3fl) 5664f)! far/351R 645-]. Amalie :0 {W [7‘35 g <55; IZ- 5 (5H)(<9#) 5549;!” far 5:5 I! 57.: 1% ft? 7778 Correspona/Ma f/au/I'm‘es {297m 59, 0)) fl/‘c‘ Q, = 01 Q, = 54791722044) 5 I28 WA) and 4), s Wf/‘(Zf/M = WW.- 124‘ a: :fiown [Ire/3W. I I Ii ll % 5.27 1:! takas you 1 min to fill your car’s fuel tank with 8.3 gal- lons 0f gasoline. What is the approximate averqgc velocity of the gasoline licxaving the: G.-i:n.—diameter nozzle at this pump? V ,4 I .z Q = (8. 8 yd/ ) _ “033/6 '70)? C , 7. $4725! 6’0 3: I and Amy/C = 72—62.. "0331c = 7770.651? 2 2 W222 sa V s M '9‘" mjfla (Wfififfiwjfln V a 62.49 gfi‘ nayjle ...
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PS6_Solutions - 531 Water enters a conical diffusing...

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