Introduction to Fluid Mechanic
Introduction to Fluid Mechanic CE 321
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This 48 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 45 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 28 Open Channel Flow Open Channel Flow the flow of a liquid in a channel ha s not completely filled A free surface exists between the fluid water and the atmosphere Main driving force is gravity Types of Open Channel Flow Uniform Flow Depth y does not vary with distance x dydxO Nonuniform Flow Dept y varies with distance x dydx 0 Gradually Varying Flow GVF Flow depth changes slowly with distance dydx ltlt 0 Rapidly Varying Flow RVF Flow depth changes rapidly with distance dydx 1 Types of Open Channel Flow UF uniform flow GVF gradually varying flow RVF rapidly varying flow Waves Examples of waves ripples on water sound waves the motion of a plucked guitar string ocean waves A disturbance is transferred from one point in space to another without any substantial movement of the material itself think of a floating cork that goes up and down on the ripples The cork experiences very little movement in the direction of propagation of the ripples Q If the material itself is not being transported then what is being transported Energy coszElt Mw Mm COED gt S w mw wv Motion The progress of the wave itself should ot be confused with the movement of water particles Wavelength Distance between two peaks w Motion Animation In deep water water does not move fonNard with a wave If we followed a single drop of water during a passing wave we would see it move in a vertical circle These vertical circles are IIIUIe obvious at the surface As depth increases their effects slowly decrease until completely disappearing for d gt k2 Wave movement I P At l j Motion of water withinC C C the wave f C C C Negligible turbulence below 112 wavelength if Largest at Surface Negligible below x 2 0 Circular Motion Longitudin and Transverse Longitudinal waves Particle displacement is parallel to the direction of wave quot Transverse Waves Animation Are Water Waves Longitudinal or Transverse FM 39 h a 39 I III I 39 II lII39I III3939I 39III IIIuIuIlgIII I39llIUIIIIIlIIII IllllIjllllllllljjjjlllll IIllIIIIIIIIIIIIIIIIIIIIIIII llllllIIIIllllllllllllllllll Answer They involve a combination of both Animation Wave Speed 3 61 Moving 7 I c wave speed end wan 1 l39 a 50 039 i i Channe width b su ace N Stationary wave 7 H x 1 3 I h H Wave Speed in Deep and Shallow Water Deep water y gt k 2 c Shallow water y lt M20 C Z 83 tanhx z x gt Smallx tanhx 1 gt Large x tanhx function tanhx z x Small x tanhx zl Large X 42 fl deep ayerthf f t f f f 34 3 J i i f Eq39104 Felts Froude Number Ratio of fluid velocity to wave speed Fr L k7 Subcritical amp Supercritw vws Subcritical Flows s v s V Disturbances eg a wave can propagate upstream Supercritical Flows Fr gt 1 V gt c Disturbances eg a wave can not propagate upstream Subcritical Critical and Supercritical IIows Subcritical F lt 1 Critical F 1 Supercritical F gt 1 Uniform Flow in 11 swim 4i 39 New mu EECUIZIII u I Many channels are designed to carry fluid at uniform depth Chezy Equation Read section 1042 Can be derived using momentum equation 39 K 9 applied to a control VOIume Rh Hydraulic Radius Ap 30 Bottom s0pe F9 67 LA 11gt j Manning s Equation From experiments it was found that n Manning s roughness parameter sm13 n depends on the surface material of the channel s wetted perimeter Manning s Equation VZERh23E 14 Q EARhZBE K 1 if SI units are used K 149 if BG units are used Critical Flow and Gravity Waves At critical flow in a channel the velocity in the channel is equal to the velocity of a gravity wave V xgy 101 Water ows at a depth of 2 f in a lOftvwide channel Determine the owrate if the ow is critical Fr 1577 w em y 27quot7 I Q Vby wfh b01 This if Fr V47 32225 802 FxJ 50 4 3 Q 8011quot oHszf 503 lz 105 Often when an earthquake shifts 21 seg ment of the ocean oor a relatively small am plitude wave of39ery long wavelength is produced Such waves go unnoticed as they move across the open ocean only when they approach the shore do they become dangerous a tsunami or tidal wave Determine the wave speed it the wave length A is 6000 ft and the ocean depth is 15000 ft From 57 04 C m l or ya 3225ooon 15000 C 75 ii 10 Watar aws in a b IUfIwida rec tangular Cha NEl with a inmate of Q 61 fth s and an upstream depth 0ny 2 ft as is shown in Fig P10 9 Determine the aw depth and the sumquotface elevation at sectinn 2 1 Z 4 L2 f a 1 g iZZI Wm 103025 Zlx2 2102f u 60g 0 60g 6 VI3939W 3 and V1 7 m39fyz 7 lids a 6 2 3 7 2622 2 2622331 O Z Uquot y L94 0551 0 Mic1 IMs no yz 774quot 0532 and 0532 V 3 s Vofe 39 r quotRal yzmjl Ify1atazl Men Frzgt ML sawof be since Mere i no quotbuIzp yewen I andlz ml tabah arHm WMWM can 0am 7 y 39 776 f and 2 77 f 0374 lt 1012 Rainwmer runoff from a 2007ft by SODfl parking lox A L L a L of S mi Delermine the pipe diamerxe if it is to be full whh a sxeady rainfall of 15 inhr 0 AR s 239 Where h I so szgfn 0000947 m D1 D A KDQ and h fD 7 From 7715 0 n 00I2 m 0 41 r Wharf r rainfall rah 152 Thus 0 200F500HI539Lr xsggs 3 Hence from EgJJ 3m 42 12 45939 avowt f4 0r D M H Lecture 27 Pipe Flow Problems4 What is the effect of flushing on flow rate from a shower How will you approach the problem 1You need to know how things are connected toilet reservoir and shower 2You need some data on pressures flow rates and losses in i e com onents 3Plug n chug v The bathroom plumbing of a building consists of 15cmdiameter copper pipes with threaded connectors as shown in Figure 2 If the gage pressure at the inlet of the s stem is 200 kPa during a shower and the toilet reservoir is full that is no flow in that branch determine the flowrate of water through the shower head b Determine the effect of flushing of the toilet on the flow rate throughthe shower head Take the loss coefficients of the shower head and the reservoir to be 7172and14respectivelyr 77 7 Vsiibiw r Had 39 KL 12 Toilet reservoir with oat Globe valve Cold 1 fully open water KL 2 KL l0 Properties of water at 20 C 0 998 kgm3 0 1002 gtlt103kgms uz 1004 x10 6m2s 0 a 10 X 1U6m The piping system of the shower alone involves 11 m of piping a tee with a KL of 09 two standard elbows with KL 09 each a fully open globe valve KL 10 and a shower head KL 12 Therefore ZKL2092gtlt091012247 Apply the energy equation between the inlet 1 and the shower head 2 O o 0 O b 2 2 10 V 10 V 1 Z h 7 I I Z I 1 P L 2 2 T V 9 V 9 amp22 21 hLgt hLz ZZ Zl 7 7 QAL W1n ms E213 my w 2 W M3 92 AQMD KLacc g x ZKL g i quot5 a Ema um Vt CD 00m mm quot 62 m I a V T7 V Re L3 Ho o rm 7 quot OLF7ltIS 39 S it 20 f E 37 0 QB 39gt we v a WEE L0Ol wxgt ka quot3 aw f 13m bog 73333 2197 61gt Ekaok z gt I LR 9e nwwg a C 39 J39 c zw 61702quot 700 Wm39wo G r gt769 m atom 39m 0L MIC r1 quwtm Win LP NnKWs yaHALMuUN quotquot Wm WM MAmkg mad1 amp 000068 MgS 39 002 235 IS I Re 44500 MW quk 72 0 0005265 4451903 577930 002176 445 1 7530 2176 QW U 40 5283 533 36 3 344 3 3436 5 3 4 lt58 4i B w 3 E Q i ogs 3 3 535 gt334 ab a WY 3amp3 AF 343 423 7 M 6 57 z ooooo M J MM 9quot 2 zloo H 293 739 amp I g L 6 Vb 9 QM 39 ZLP39TM 2 039 O 39W 2mg I l V V 33L 3 9quot c 7 Q M amp LE 512 1 9 M 2 TV 9 TV D MA 4 ome3 3 TILomCMB 39 gt I UV 395 Q1 403901QM 3 r 39 mc v gt Re I r CO M L Loolhq bf WOW 30 f S g 213 V3o01 i quotOOLHd oikk L9LD M H No M A Qq f 37 mow 11 Qe C39JA M m Lzrzobg 39m amp if 37 039 OK 29ng 2 23 Jrf39f 39m Q3 I 31001Cw Reg J12 395 f l M I f l91 O 000 s Q2 S 3 W mA MYWm W 06 M 01W m l 033 i 4 0L L S x 863 The three water lled tanks shown in Fig 13863 we connected by pipes as indicaLcd If minor losses are neglected detcnnine the owrate in each pipe Elevation 60 m Elevation 20 m Elevation 0 Axum f7e uid ows from f g 3 and Ah 0 mm 0 z 01 W3 or galI701 v zogmfvz Lgmoam f m V Q64 V2 054 V3 0 For uid flowing from4MB whlh 70330 and quLQWJ 2quot25 gIIEY gbl Vf or 2 z 29 V1 V2 00 200m 1 60m 20 00151 010 Mia5 oozoquotooem 2 q 8 m1 Hence 5 9 0 3 529 V 2 255 v12 Simiary for fluid owing from x4 7 06 will7 ffc 0 and 944 0 2 1 If r amp Ya 3 61 5629 3032 0quot z 1 2 0 I 200 v m Va 6 m 0 05 039 0020 008 Elevation 60 m A Bevation 20 m 2352 Elevation 0 I Hence L6052 72 510Z Sove 57512 and 3 for VI V1 quotth F m gsHand 57 604519046 1V1 v315a 5 or 958 Vz v31 am 4quot swim 571 rm 713 50 40 5J0 142 25514quot or l2 lm3 734 77m rm 5p sounds 841W 02 958 0 7773 can be Sim ed f0 2 K712K51 Z 9439 03 6 l 7 V31 Square be 546 and rearrange 0 give V 3 V32 6725 0 wzich can be saltcal by H79 quadrant 1 formula 1 0 give 2 443 l9 631 4q25 3977 or 7136 77 V3 343 or V3 230
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