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# Introduction to Fluid Mechanic CE 321

MSU

GPA 3.85

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This 31 page Class Notes was uploaded by Jolie Shields on Saturday September 19, 2015. The Class Notes belongs to CE 321 at Michigan State University taught by Roger Wallace in Fall. Since its upload, it has received 46 views. For similar materials see /class/207369/ce-321-michigan-state-university in Civil Engineering at Michigan State University.

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Date Created: 09/19/15

Lecture 23 Major amp Minor Losses Turbulent Flows Motivation To determine the head loss that occurs a pipe so the energy equation can be used in the analysis 0 r we m wwwms Two Types of Losses Major Losses Head loss due to viscous effects Minor Losses Head loss in various pipe components h L minor hLzh L major Roughness The roughness of the internal surfaces of pipe in contact with the fluid is important If the average height of roughness is s then sD is an important dimensionless parameter This means that while comparing flow through pipes similarity we have to scale roughness as well Roughness Values I TABLE 81 Equivalent Roughness for New Pipes from Moody Ref 2 and Colebrook Ref 3 Equivalent Roughness s Pipe Feet Millimeters Riveted steel 0003 003 09 90 Concrete 0001 001 03 30 Wood stave 00006 0003 018 09 Cast iron 000085 026 Galvanized iron 00005 015 Commercial steel or wrought iron 000015 0045 Drawn tubing 0000005 00015 Plastic glass 00 smooth 00 smooth Q Why new pipes Shear Stress Difference between laminar amp turbulent TIOWS Dimensional Analysis APfV D8p k7r 3 k r4 2 Ap pVD i IAPD PVDDEJ lpyz D u D EpV2e D 2 ar FrictionFactorf e V2 u 7D D 2 8 f 001675 Energy Equation 2 2 ILZ1 ampVizzhL 7 2g 7 2g D1D2 gt V1 2V2 2122 171 172 ZAPZWL IDPQWWWH J 7 l a 2 r J The Moody Diagram The functional dependence of the friction factorfon the Reynolds number Re and the relative roughness comes from experiments This data is summarized in the Moody diagram It is not exact accurate to within 15 of data Wholly turbmem flow Laminar ow Smooth Transmon range 103 10 105 10E 10 Rt M u Friction factor f 64 f Re f 08 D Fully developedturbulent ow Large Re f 0Re 8DIlVlO161 at6 valuesof Re gD 251 1 F 2Olog 37 Re7 Q What is the problem with using this relation A fappears on both sides Needs iteration J lCoIebrook Relation l Friction factor f An explicit relation for f was proposed by Haaland in 1983 1 69 all I F E 18 log ng Re 37 Gives results within 2 of the Colebrook relation A good initial guess for an iterative scheme Pipe Flow Examples 3 Types of problems Type 1 Flow rate Q is given Find pressure drop or head loss Type 2 Pressure drop or head loss is given Find flow rate Q Type 3 Pressure drop and flow rate are given Find I i e diameter D Explicit Relations 2L D l9 hL 107Q 5 1n 5 462 W gD 37D Q Z 10 ltgDlt10 Z 3000lt Relt3x108 5 l5 2 l5 Q 70965 gD ht 1n 5 33917 L Re gt 2000 L 37D gD h 2 475 52 394 6 Z D066 51ng VQ94L 10 ltgDlt10 ghL ghL 5000ltRelt3x108 P K Swumcc and A K Juin quotExplicil Equations I39m Pi pchlnw Problemsquot Jouma aft16 Hvdmulic Division ASCE 102110 HYS May 1976 pp 1574364 Minor Losses Losses due to components such as valves bends tees etc expressed in K hL minor APL dimensionless form L 2 2 1 2 V Measured V mg 0 Expressed as a loss 1 2 coefficient KL APL K L E pV Changes in pipe diameter introduce losses as do inlets outlets KL depends on the eometr of the com onent Why Minor Losses Because the fullydeveloped flow calculation for the friction factor f does not take into account the losses due to sudden changes in pipe diameter etc Minor Losses Pipe section wnh Valve I 1L KL 2 V lt 4 V garava hL Lpg Pipe scclion without valve V PI P2p1pe J APL 1P1 7 Pym 4P1 425 Valves allow us to change head losses until the desired flow rate is obtained A globe f 33 valve Vl l gt V2 1 Angle Valve Constric on v2 vi Vuunqncnon gt V1 I Globe V Gate Valve Minor depend on gametw KLO8 KEG2 KLOO4 Cc Entrance Loss Effect of Rounding Inlet Edge on KL 05 it A O 3 212 11 Cl J was 11 015 120 025 Sudden Change in Area 1 Sudden Change in Area 2 J 3 l I7 L 0 39 A zg 039 K 0 D2 04 on a Exit Flow Conditions Energy Equation Including All Losses Major amp Minor 2 2 VLZ2amp4 Q4 Z211Lmajor11Lminor 7 2g 7 2g 2 2 2 2 zlz zzf LZKLL 7 2g 7 2g D2g 2 179 819 During a heavy rainstorm water from a parking lot completely fills an l8 in diameter smooth concrete storm sewer If the flowrate is 10 1133 determine the pressure drop in a loll ft horizontal section of the pipe Repeat the problem if there is a 2 H change in elevation of the pipe per 100 ft of its length 331 71 sbldtign In generalJ 21 21 4 g fi 22 14 EV 0 wvere Z my 339 pg and 0 I0 I V V VZ T W 5 77m 7 I git63955 A m biz39pv m g 000 f I 4 M7 m For a smm awards pipe 5 0 X La and Re 71 4 53956g j Z 92 xt75 2Ix039 5 77m from H 8J0 0085 5817 Solution 1 WW 24 1 gives la 1 f 4331 2 11 H A WWW55 Egg 554 3s2z m ozags b WM W 1257 Z 279 so Mm 2 4 65sz 432H4yf z 0155 3 Esi C WW7 fow dawn7i ZL Z39H 0 Mint Af52 s ZM7ZJ 7 0255 0005i 820 Carbon dioxide at a temperalure 0f0 C and a pressure of 600 kPa abs ows through a horizontal 40mmdiameter pipe with an avcrage velocity of 2 ms Determine the friction factor if the pressure drop is 235 NmZ per IO m length of pipe 1 D For a horizonfal Pipe A 7 D 35912quot f Z We 600X035N 2 k M63 1 P 7 I683 273 K m3 m 4 w as N 2 00 m 2 E 0009 115310m 25m1 Determine the pressure drop per IOOm length of horizontal new 020mdiameter cast iron water pipe when the average velociiy is 17 ms V2 Z l l 1 51 2 47 2 zz1fb or 03 V 14 V 731 I 1400 D 02m 2W 7m 1 40 V0 701 Wirzr 439 F39s Fipvz F 1Lsz ff19 pzl39oe is barmanat 3 But from 7amp56 8 g M4 3x0 3 02m D Hm 07 Also Re 3 3m05 so Mall fryn 8 IO J f 0022 7771 I 1 y A oozz2 4977 27 59x0 5 741 36 When water ows from the tank shown in Fig P836 the water depth in the tank as a function of time is as indicated Determine the cross sectional area of the tank The total length of the t6in diameter pipe is 20 ft and the fricu tion factor is 003 The loss coefficients are 050 for the en trancu LS for each elbow and 10 for the valve 100 200 300 I S Z 4 Zz 4 zz 5 Ware 4 aw 210 2 324 V517 11 zq d 6L Fif Kl xi 05 a lg TAM V1 l V 2 5 127 fa kzzz Cmsider 7 faw MM f7 90 47 Z 6457 elm L 6459 MB 2019 l Wm or V 305 1 50 717471 Q qu H13M it 0 Mfg

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