EML4304C HW2 solution

EML4304C HW2 solution - EML4304C Concept questions: 8-12C....

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Unformatted text preview: EML4304C Concept questions: 8-12C. How does the wall shear stress 1w vary along the flow direction in the fully developed region in (a) laminar flow and (b) turbulent flow? The can” shear stress (TM, VCMQiWS werml‘ «level We “NJ ollreclwn .‘A H»: Lila alewalofd refit” 4L);- lmirk lawn-ware QWOL ‘l‘wf‘AWl9/%+ ‘Clow. 8-13C. What fluid property is responsible for the development of the velocity boundary layer? For what kinds of fluids will there be no velocity boundary layer in a pipe? The {ilwa V;S(9Cll3 l5 {Cgrms;L\e fir +M Xeyqla-PWLQH ‘7; film valdtila luauALILVJ 1L1zr, There MN be M we;th ‘0U(€f (5 Jrl/xfi ~Cla~d l5 {AI/Chat , 8-l4C. In the fully developed region of flow in a circular pipe, will the velocity profile change in the flow direction? Mo 8-15C. How is the friction factor for flow in a pipe related to the pressure loss? How is the pressure loss related to the pumping power requirement for a given mass flow rate? BOW ayb rra For-lLH/wbl , 2 Milka/£31, 1 WV“? 2 NAP“ “ /0 8—16C. Someone claims that the shear stress at the center of a circular pipe during fully developed laminar flow is zero. Do you agree with this claim? Explain. \1 a 5 r» i u. lt‘mv 5: cc, 0&3“- ; Mu: ruler (T ML atp ’\ 0E ( 0 at, i / Gettier : 0 8-l7C. Someone claims that the shear stress at the center of a circular pipe during fully developed turbulent flow is zero. Do you agree with this claim? Explain. «as We ska» sluss off W9. SM‘Q-Lcea Di 0»)?th ohm“: tuna oleuelprxnf hamlet/d Elm.) is mud/hum 5mg w ska.» aims; :3 Prafarf.wl( +0 ‘HAZ Velociillr j/dxje‘tf LQLLWC/l/k i5 Maxiiflum m‘l' “Hus, +«AQ SWI‘CILP? / 8-18C. Consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the head loss will (a) double, (b) more than double, (0) less than double, (d) reduce by half, or (e) remain constant. 3‘wcer. \AL ; E12352 ) dpkhlsfi ‘HAQ [walla ciao.th {fl/re, M US; 8-l9C. Someone claims that the volume flow rate in a circular pipe with laminar flow can be determined by measuring the velocity at the centerline in the fully developed region, multiplying it by the cross-sectional area, and dividing the result by 2. Do you agree? Explain. %e5~ ‘V‘ : Vm‘A . FW 37/” (6“18/ (‘L/VHX 7- ll/du3. “m” OCCU§ ot'i confer Ive, ’ ’;\/MQ\{ V :A, 8-20C. Someone claims that the average velocity in a circular pipe in fully developed laminar flow can be determined by simply measuring the velocity at R/2 (midway between the wall surface and the centerline). Do you agree? Explain. N07. “(4'): ZVWB (\—%\ I [ref “0-) ; VAUS aqafi r. vyfi-riv/fll-Lét) Lzz-ci -'-_fi HER ) :2 R7. ) 2 K7. ) TlMA/S Vow/3 ocgufs ml J}: L . 8-21C. Consider fully developed laminar flow in a circular pipe. If the diameter of the pipe is reduced by half while the flow rate and the pipe length are held constant, the head loss will (a) double, (b) triple, (c) quadruple, (d) increase by a factor of 8, or (e) increaser by a factor of 16. k“; LEV”??? lLZVTLE: 14" D 15 (dwelt, . I h i to. [War HM, 9:96; .1 V :wfim J 3 k M, lL Rik 1 @i L: V‘ g at “LVL ti I will .acremse in? L RCDE' vb ghnglaLfi-V ' L“: j i 11' Bl“ M1 1 ’ P 3 M ‘ 3/10 8-22C. What is the physical mechanism that causes the friction factor to be higher in turbulent flow? IA thrloe‘\€n‘l' New) fl 3; flag ikadmlv am;45 4kg +0 CAMwL£<£ Wilkins? ‘Hml (tuft? ‘llne £r:clfim father” +0 lflQ lQV‘TXV‘ 8-23C. What is turbulent viscosity? What is it caused by? Tamale/mt Vieoojhlng la; :5 Cwsex 95 ‘Ig/‘(ItclP/M'l‘ wares; am! ML. accau/ulé flat" wwwflm lrczmjfd/‘l lurblel 2106(85. it {s exWKfiycaé k5 (Ti: 3 ~/0 UJVI =At if whgre (I E, He main Vqlqe g-C V5lo(.‘%] [‘4 HM: “glow cllreclfim anal» Uv/ ckqu V’ “Ye. .{le‘I‘M‘l‘qu Wéomekfi a-p V&[a(.¥7‘ 8-24C. The head loss for a certain circular pipe is given by 92 hL=O.0826fl;35— where f is the friction factor (dimensionless), L is the pipe length, ‘6’ is the volumetric flow rate, and D is the pipe diameter. Determine if the 0.0826 is a dimensional or dimensionless constant. Is this equation dimensionally homogeneous as it stands? in MS (“uh of +kere-Fgre) RHS Musl’ MS tank's a-l‘ ClCck.‘ I46 “’5 ‘1 '3 s 1 -5 _ [L] : Lomé] [A [L1 : L T : LiT 1 SWLQ- LH5 "3 Mi ollm;m01\¢llq CQ’KSZE'l‘OMT EH9} 1062.9 Mus‘l have Jilmwsims] w QKP/‘e $5.9,A \3 Vlo‘l‘ axiwtwsl‘d‘flcf“? M‘LOCSWEO‘LJ, 8-25C. Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the flow rate is held constant, how will the head loss change? __... ’— ’- L D Re. D D 11* fab; it a :5 mlmd t3 a m l... W ‘09 “Bird ‘13 Vii \fi-_QL—.!:,Q;LL~..V7’ ,CjiuLvy_ szaLV / 8-26C. How is the head loss related to pressure loss? For a given fluid, explain how you would convert head loss to pressure loss? infill r: Allfliku 8-27C. Consider laminar flow of air in a circular pipe with perfectly smooth surfaces. Do you think the friction factor for this flow will be zero? Explain. No, Due +0 46* uo_g(.'|(\ ~HM241 will be A VC\"CA-{’8 if’wgum‘l‘ ad “We WOLHJWWOQ~ HMS a. shear" gifts; in 'ku; ‘plw I 8-28C. Explain why the friction factor is independent of the Reynolds number at very large Reynolds numbers. [Hr Vevg lapogg keTml£5 “WIMBGP5) 15 Marla cud the @Lcllm Qaciwr is Mia/JonaQewi" a“? ‘Huz/ Raulnollé number, Tim ls Lac/Mae, the lei/tame?) 0‘? HAL law/MIMI failvuler decreases w'd’k Marmara] Mm”; kULMlOQCJ 01ch Hr Mum as 50 Hm Hog? HM, garfima Moufluxeés wpalrwks ml?» fie flora, The viscom ewe-(Acc’i‘s M HMS Cage am ff‘ovhccmi M flee Wtch flaw Vim-“AV la a WWW WW: «mm M m cm: Mm ) 0% % law/«Mar éw‘o lat/(12$ \«LWKS Reflux;th Other problems 1. (8-37, C&C) Consider an air solar collector that is 1 m wide and 5 m long and has a constant spacing of 3 cm between the glass cover and the collector plate. Air flows at an average temperature of 45 °C at a rate of 0.15 m3/s through the l-m-wide edge of the collector along the 5-m-long passageway. Disregarding the entrance and roughness effects, determine the pressure drop in the collector. Answer: 32.3 Pa -5 “V: 1'75flxm “47/5 tog/UDC _ f- not my“) D 2 _ m l “ it; : L” 03 y M Jaojgzsm 1x,(}X/{.+2KI7A Re: VB ' .13 Z A AZ}: #:%%1 // X49552.54 :M/Hs .fiwm m5wo'5é: Waco 15 id [WW/ml so 'PfiAu-J‘A author [5 «paid from MaopLJ OLIu‘g/WI 1.017 Ffiwm Calahfook (36". I 017/ AfzwflL/“figfi- Luvlz '2 L t ~¥vfi~»,e 3y;— 2 L DzkALB'Qpé'HawL-ésL—f—L ZHLU'L AFL : .017lx5X140filI? r"5m3/5‘)Z .05925/ M“; QWV M35212; M, :15 gr PM M/‘f' Jfiw’ $3: ) APL : 32.25 FKJ 2. (8-41, C&C) Air enters a 7-m-long section of a rectangular duct of cross section 15 cm x 20 cm made of commercial steel at 1 atm and 35 °C at an average velocity of 7 m/s. Disregarding the entrance effects, determine the fan power needed to overcome the pressure losses in this section of the duct. Answer: 4.9W d/y/‘VW 20 ——~— A 7MB f- LHS fi/fi Uh Ml 35°C 0L: l’?15KlO-5 5 .l 7“ I “‘l‘“ ‘ lmt 7/2 i.6§§x<l05 6: 05‘5 mm bk; L‘AC m... 7;”: Lli. X.|SM ’ 0/[7/7M ulm + IKJSM Re: VDL‘ 7m x 7 r 7: /5 HZ“: = 72,4625 M/g \& 1 ’0 M {b u 1 2,425x10’? [71+me PW celelfloch; éfiua‘lz‘“) ‘(3 f .0203 Ll 7. M2451. 0 [<3 ‘ , _ ' "l 2. fl 9 I n “i WV‘M‘OX: l [(1552x lMx/ISMXIUZO5‘QXM ijM/SBA M [I‘M-I” "T 2/ wm- Warsaw fiat/r :21 fl 5/ JV MM ,1? W = Lt m w) 4/10 7/I0 3. (8-46, C&C) Glycerin at 40 °C with p = 1252 kg/m3 and ,u = 0.27 kg/m-s is flowing through a “he San—Wm horizontal smooth pipe with an average velocfi'WT‘fij m/s. Determine the pressuréxiglgpjper 10 m of the pipe. Ra: Pig 1 [mrzfixafiflénwfl I .2?! 93/5”? RNA.) LG i02er mu) 5:3 {1: aki/tC—y ’- -07%€'7 Z Foi‘lbm) :Q'Lsk f 2 21: (315m/Sjl W m I43 L Ar 110,103 MA ’61?“ )6 (3% 4. Water flows in a horizontal constant-area pipe. The pipe diameter is 40 mm and the average flow speed is 2.0 m/s. At the pipe inlet the gage pressure is 450 kPa, and the outlet is at atmospheric pressure. a) Determine the head loss in the pipe. Frm 61.46111 whom, him»? 1 Mai/[4,711 : 0 (Em f‘ol> , 1 f3 ""+%I“/T;)"*'°‘zvl~_+gz. +l\i Mét=0 ) “‘6' tr? “ié’ 84°C., f: m: Md M to? Q‘1 {1: “L: _ “50% 10009.: )0“ 5w ikh : b) If the pipe is now aligned so that the outlet is 30 m above the inlet, what will the inlet pressure need to be to maintain the same flow rate? J.“ ‘lkl‘l’ 0)) lL ‘5 £RQ ‘l'n vl5w3:¥o_:¥ 'IS QMG) hL JJVL‘i’o VlScaSX‘i‘g L5 5M€r LL : P41; '22 P3 R'Pz : QTUEL>IA§ P7,:0 f. 3 (“Mi/a +30%)x llggx‘fll a i": EEK m i” W M [703% Ssz H C) If the pipe is now alig®o that the outlet is 30 m below the inlet, what will the inlet pressure need to be to maintain the same flow rate? For Y110 7‘ : (Lg at)“ :05. 7éy~3flfl)* tfitfiflmg, M 24.? fit if tea/id my zRT lSéSkQL ‘ d) How much lower than the inlet must the outlet be so that the same flow rate is maintained if both ends of the pipe are at atmospheric pressure? \>\‘—‘?1 Msz $1: ’kL ‘7/0 5. At the inlet to a constant-diameter section of the Alaskan pipeline, the pressure is 8.5 MPa and the elevation is 45m; at the outlet the elevation is 115 m. The head loss in this section of pipe is 700 m. Calculate the outlet pressure. Use 0.9 for the specific gravity of crude oil. f? ~‘l/7Ma " .‘Tt CH? “3&3 Fm emern when? kfuuiqaz Limit” " °<«Ul : “2U; L+‘Z:E’L/+2Z+L\L m ' M. lz‘ mefi/Jg ~%z/33~LL/>3 V1: iDt‘IC‘EI'ZZ’kLB/A‘j M P7, 2 92'5le + um — //5,4—700A x m WK x7.?lZ{-MMMPQ. < > M OJ 106% EETs SSQUAEE ears 5 58mm: 5 UARE EETS 5 50 SH 00 SH 00 SH \sofuwbn: App/5 the end/:99 eyaafién 767* .529:wa xécofipmas}飢. pr)“ from. Compmhfig Qua-2:560 : _i -2 (g’i- “11;: 4-33..) ‘( HOV-235 * gagf'x’lr 77am @‘MW :5» or Plus and Gait/m: @uaéi-wadvj flow 01" watt/- 793m {amt fihown. AO“ 0-5. 1.11:2 :A2. T Raid: (0.) Fiona rate: at {-40.57‘62/11" show/3, (A) Howl Could deaf-rm x'mpm ued :9 I L Assmp-h'aas: (I) 19, 279,: spam “'3. Eur mm Cow/flaw, {2,4, = 12,4le 0;: L7: (gatmd p: .afz‘szzfiflag a =- Vzm = wager-d prim, . .... .. ._ (2.) .. addxhg a dfifi‘msa‘. - — - *a. V -—Vz V: V: %+?;,—_3_+)<__ u(r+:<)__ i Frobfdm 8.39 A 1 =‘ {517 17 L’ZJ'Negfcuf- mam}; In short- tube; L030 (.5) Rem/2+ eflflanoc, K_‘=0.7z (Tab/(£2) (‘0 Unrfi'rm 'flaau 41‘: each 4:617:3an IX =1 7. 2 Z _. 235, f‘f'K “(Axmifil 10.9 a"; /.s I “if: 52—" “ notaring "' a 793 .s _ 149m} 1" fwtal #318! . (14.29pm) a How; EML4304C 1. Water flows through a 60 mm diameter tube that suddenly contracts to 30 mm diameter. The pressure drop across the contraction is 3.4 kPa. Determine the volume flow rate. M59, $31) 5LLL£€M coni’rocc+:m 40 $61. KL 7. E _ BOW-MB : ’15 D1 _ 667M.» KL‘ (27‘ LL: KL whiz (/ is W530? JR deiacii-a in {MN ’{rfi Wm WM r MHZ Li£+24kflagé+£é+fl+wiuth P3 9.3 2 V22 “VI-Z PF PL 2 + L“. M 21 V1 Ve\0C~i~?’ .. («112‘ Piffl ' V2,?» Va'ocf‘izl M {MN 1 Z ill—LE} " Vfl‘ +' KL [/12 W 22 “3}” 2/7 ‘Q: 0.0m “75‘ 2. A pipe friction experiment is to be designed, using water, to reach a Reynolds number of 125,000. The system will use 2.5 inch smooth PVC pipe from a constant-head tank to the flow bench and 50 ft of smooth 1 inch PVC line mounted horizontally for the test section. The water level in the constant-head tank is 2.0 ft above the entrance to the 2.5 in PVC line which is connected to the 1 inch line. Determine the required average speed of the water in the 1 inch pipe. Assume T = 70°F. Calculate the pressure difference expected between taps 12 feet apart in the horizontal test section. Please work this problem using English units. Do not convert to SI. Comment on the feasibility of using a constant-head tank. A «NW2 1: 1,0 41 + i T Pa 1 0.500 0 ‘1) Ra ; /’ V5918. D 1L f: “Jum- ‘F’ i -‘1 4“ 6.556 XIO $3 dip—2f Vane: K” ‘W ll Z Li M ° 4&6‘1'4; FW 49de fiiclrm ) leg 0’0l75 P‘ I? 4- : L+ . +— R+qulx k/ldl rat with (L3 FefiUfikfilkCW ~ 0‘? Wk ?fe,6{u,vc, at} +0? 0-? '- fFQ-55u"ev oi EXH’ : POL ASSW is lde‘vZ, 5y vetacl‘xa ‘m “-nr\’§‘0~0. (A5 a VA‘ .‘-V2" «_ a, [D 7" v 0—: we. ’7. 22"— b—E+Z+L 13 13, L L+D+t\ _,‘.\ leZDZL' 75-13% v Emmw ¥/ 7 uh. w:- IL 2— % 41‘” a? i a L\ + 24+- Mzfi = 1/}: V033" j+ «EL 5wqu—2 +QQ€L M23731 2.6 \D‘ 27, 239+ 0‘ Z} z Z 2. ‘1 . A \/ |/\{\—~ %‘ :— [l‘ '5‘. Téop'élzl’ZH 7,, 1 1“: ML [1- WOW/11’2"“ I}? 3_ fight /Dz>‘+ {4' Dr 17” -'})2 \1 a =- W13: WW“ ’2‘” (HM? E3, we 1 ’H1<O‘ >r Mili- “gt M 3:» 1- VRQ: PV‘Dy : VLLP'DLBLZQLI?) fig/X L1: w a ’ Kai ’ A, éuJUQK‘O R“ t 3617??? '79 4“ 3 ,0117 \n ; [l {$753.5 éOQx 491763 ' 1 H“ hwwflsfl l _ g x,ol|7 (IS-"lgmlz’g?" M 2.x. 51-1 M, qm 19+ ff: mm; he, Jr.” M 3. A hydraulic press is powered by a remote high-pressure pump. The gage pressure at the pump outlet is 22 MFA, whereas the pressure required for the press is 21 MPa (gage), at a flow rate of 0.029 m3/min. The press and pump are connected by 55 In of stainless steel tubing. The fluid is SAE 10W oil at 40 C. Determine the minimum tubing diameter that may be used. ) 7MP ‘._mmwm_m.ri:rm——m_w,_::fim :th Efl€'&a 86.1 P; w? hr + ,+£1 >3 ELAN/L +"(n-f +‘n r3 5-? f f2) éd/z M L OZVI;dV1—} %.:‘ f L V2 r T LIV «Q: Riki DB 172 \ s/3 T Y W W7 V71 3" Rem“ - wzifilb;m Re PM' fl D, amp SO ” n2. ‘ oomwsm 2,5lx3/6X‘0 Mi“7r"b ._ T N W” Zyflm 3 ’I D Lt x,72x77?r)ixv2_zlfiy 19513302 ,05!Y?E3 “2/ 606/ M m5 , l)(|.~,M1 D m __J/«~ 2143.0 5 W " + MONO“ / m D ms Ms . i I r - §.fR ms”? EI-Vrk a) Cu CM °L+O /A “M Q; (2—: My So/wArJ—rx.» ‘Qfi ‘hrbdw‘l" flat»). “‘3 [M‘Aa’ ‘ 7 £1 10511‘1135. 91 : WNW; 1W7?” ' Ra w M D‘i : MMTFMJ _/_ : if, x Zémézfi ‘T Go! 5% It! ‘A. M w M W W (05207266 WNW/w»qu "Y‘ D: ~0"/W\ ‘ W “W1 acucmwwhwr ...
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EML4304C HW2 solution - EML4304C Concept questions: 8-12C....

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