Dinamika fluidov. Laminarni in turbulentni tok Viskoznost tekočin Faktor trenja h f
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1 inamika luidov Laminarni in turbulentni tok Viskoznost tekočin Faktor trenja h 1
2 Energijska bilanca: Celokupna energijska bilanca procesa: W 1 + U 1 + K 1 = W + U + K F + M + T Bernoulijeva enačba Enačba velja če privzamemo da: Je tekočina nestisljiva in neviskozna P 1 1 v1 gh1 P v gh 1 Ni energetskih izgub zaradi trenja med tekočino in steno cevi. Ni prenosa toplotne energije na meji med tekočino in steno cevi (toplotne izgube, gretje ali hlajenje). V sistemu ni cevi ni mehanskih črpalk. Tok tekočine je laminaren in stacionaren Laminarni tok: Plastovito gibanje tekočine Hitrostna porazdelitev drsenja plasti tekočine v cevi:
3 Bernoulli Pressure Lowering The linear drop in luid pressure is according to Poiseuille's law, but the constriction produces and extra drop in pressure according to the Bernoulli Principle. višine. 3
4 Viskoznost tekočine: Leta 1678, je Isaac Newton podal Newtonov zakon, ki pravi,: The resistance which arises rom the lack o slipperiness o the parts o a liquid, other things being equal, is proportional to the velocity with which the parts o the liquid are separated rom one another. Odpor tekočine proti toku je pri enostavnem strigu linearno sorazmeren hitrosti strižnega toka oziroma hitrosti strižne deormacije. Faktor proporcionalnosti je v tem primeru»newtonska«viskoznost (). 4
5 Enostavni strig viskoznost tekočine - osnovne deinicije enostavni strig: F x v x A h tekočina y strižna napetost: F strižna deormacija: yx F A F viskoz. A strižna hitrost: d v dt dv y dx F v x v F d dt 5
6 Enostavni strig viskoznost tekočine - osnovne deinicije σ xx yx zx xy yy zy xz yz zz xx yx zx xy yy zy xz yz zz yx xy d dt 6
7 Ni energetskih izgub zaradi viskoznega trenja med tokom Padec tlaka v cevi pri toku zaradi viskoznega trenja F v F v F v F v F v Poiseuille F v A v e R P R 8 L R 8 4 P L L A=r Poiseuille jeva zveza se zelo dobo ujema z eksperimentalnimi podatki za tekočine z newtonskim bnašanjem (Newtonske tekočine), ko tekočina teče plastovito laminarno, torej ni turbulence. 7
8 Energijska bilanca: Celokupna energijska bilanca procesa: W 1 + U 1 + K 1 = W + U + K F + M + T 1 Bernoulijeva enačba: 1 P1 v1 gh1 P v gh Hagen-Poiseuille: Newtonov zakon: F v A v e R P R 8 L R 8 4 P L Strižna napetost: F viskoz. A 8
9 Pomen Poiseuille zveze: F v A v e R P R 8 L 4 R 8 P L F v Pretok tekočine je 100 cm 3 /s. S spremembo parametrov enačbe dobimo: Če x povečamo, bo pretok: dolžino L 50 cm 3 /s viskoznost 50 cm 3 /s tlak P 00 cm 3 /s polmer cevi R 1600 cm 3 /s Pretok krvi skozi žile: že majhna zožitev žile zaradi npr. poapnenja ima presenetljiv vpliv na pretočnost žil: R 1 = 0.8R R 1 = 0.5R R 1 = 0.R 9
10 Laminarni in turbulentni tok Za razumevanje razlike proučimo Reynoldsov eksperiment iz leta 1883 Linearna hitrost pretok tekočine Pretok narašča hitrost narašča Laminarni tok: tekočina teče - struja v plasteh Prehodno področje : plastoviti tok postane valovit Turbulentni tok: tekočina ne teče v plasteh ampak se zaradi vrtincev giblje tudi v radialni smeri. 10
11 Osborne Reynolds,
12 V laminarnem toku je hitrostna komponenta strujanja le v smeri vzdolž plasti vzdolž cevi. V turbulentnem toku je prevladujoča hitrostna komponenta strujana vzdolž cevi, vendar se pojavi tudi radialna komponenta hitrosti. einirano na osnovi dimenzijske analize Linearna hitrost pretok tekočine Reynolds je ugotovil, da lahko nestabilnost toka napovemo z razmerjem med hitrostjo toka in viskoznimi silami, nestabilnosti toka so odvisne od razmerja med kinetično energijo in viskoznim trenjem tekočine: Reynoldsovo število brez-dimenzijsko število: pospešek dušenje Re v vztrajnostne sile viskozne sile v v Laminar Turbulent Reynolds-ov apparatus Prehod iz laminarnega v turbulentno strujanje, ko je Re število okoli 000 1
13 Linearna hitrost pretok tekočine Laminarni in turbulentni tokovi Nastanek turbulence: Kinetične sile oz. vztrajnost sili tekočino v smer toka. Ko le te postanejo prevelike. Viskozne sile delujejo v smeri zaustavitve toka. Plastoviti tok tekočine daje zaradi viskoznega trenja prevelik upor. Zato nastane vrtinčast tok, posledica vrtinčenja je turbulenca. Re > < Re < 4000 Re < 300 V ceveh je za prehod iz laminarnega strujanja v prehodno območje kritično Reynoldsovo števiko okoli Re crit = 300. Če eksperiment izvajamo zelo pazljivo v popolnoma gladkih ceveh lahko tekočina teče pri pogojih laminarnega toka, tudi do
14 Primer: Pretok mleka v cevi: Mleko teče po cevi premera.5 cm pri 1 C. Ali je strujanje laminarno ali turbulentno, če je pretok mleka 0.1 cm 3 /min. Iz priročnika dobimo podatke: = 109 kg/m 3 =.1 cp Reynoldsovo število =.1 cp = Ns/m (Pa.s) F V = 0.1 cm 3 /min = 10-3 m 3 /s A = /4 = m Kontinuitetna enačba: v = F V /A = 4.1 m/s Re v 0.05 m m / s 3 Ns / m kg / m Re število je brez dimenzijsko! Pri računanju moramo paziti, da uporabimo enotni sistem enot. 14
15 Primer: imenzioniranje toplotnega menjalnika za hlajenje mleka Mleko se v toplotnem menjalniku ohladi iz 0 C na 3. Pretok mleka skozi menjalnik je 10 t/h. Toplotni menjalnik se sestoji iz cevi premera 4 cm. Izračunaj koliko cevi je treba vgraditi v menjalnik, da bo pretok zagotavljal Re število F m =10 t/h F m =10 t/h Iz priročnika poiščemo snovne mleka: = Pa.s = kgs -1 m -1 =1030 kg/m 3 masni pretok F m = 10t/h =10000/3600 =.8 kg/s presek cevi: A = x 0.04 /4 = m volumski pretok F v = F m / = (.8kg/s) /(1030 kg/m 3 ) = m 3 /s 3 Re kg m m povprečna hitrost toka skozi eno cev: v s m kg m s Re = x v x / = (1030 x v x 0.04)/.1 x10-3 = 4000 v = 0.1m/s volumski pretok skozi eno cev: F v = A x v = m x 0.1 m/s = 0.6 x10-3 m 3 /s Število cevi = zahtevani pretok mleka/pretok mleka skozi eno cev:.7 x10-3 /0.6 x10-3 = 11 15
16 Laminaren tok V tehnoloških procesih se tekočine med rezervoarji in reaktorji transportirajo (pretakajo) po ceveh Tlačne izgube v zaprtih cevovodih: Na osnovi Bernoullijeve zveze ugotovimo, da je tlačna razlika pri stacionarnih tokovnih pogojih v zaprtih ceveh odvisna od spremembe višine in spremembe hitrosti toka zaradi razlike v preseku cevi. Poiseulle tekočine so viskozne, notranje tekočinsko trenje in trenje tekočine ob površino cevi povzročata dodaten odpor proti toku in padec tlaka v ceveh: viskozni vplivi Celokupna energijska bilanca procesa: W 1 + U 1 + K 1 = W + U + K F + M + T Viskozni vplivi: Važnejši vplivi: padec tlaka v smeri roka pri enakem preseku cevi Manjši vplivi: padec tlaka pri toku skozi ventile, razcepe, kolena in drugih sprememb preseka cevi. 16
17 Laminaren tok Viskozni vplivi: Važnejši vplivi: padec tlaka v smeri toka pri enakem preseku cevi Energijska bilanca stacionarnega toka ne-stisljive tekočine po ravni gladki cevi Energetske izgube zaradi viskoznega trenja 1 1 P1 v1 gh1 P v gh P P Trenje tekočine ob steni je sorazmerno z dinamičnim tlakom. P F A tr 1 ' 1 v F tr. sila trenja tekočine ob steno cevi A 1. površina cevi c, aktor trenja riction actor A 1 =rl A=r Teoretično in eksperimentalno je dokazano, da so energetske izgube zaradi trenja tekočine ob steno cevi odvisne od Re števila 17
18 Laminaren tok aktor trenja riction actor Viskozni vplivi: Važnejši vplivi: padec tlaka v smeri roka pri enakem preseku cevi Energijska bilanca: Ravnotežje elementa tekočine na dolžini cevi dl tok tekočine P+dP F P tok tekočine dl Sila zaradi padca tlaka: Sila zaradi trenja ob površino cevi: dp x A = F = dp x /4 F tr. 1 v ' dl A 1 = dl A=r 4 dp dp 4 ' ' 1 1 v v dl dl P 4 ' v L 18
19 Laminaren tok in aktor trenja riction actor Viskozni vplivi: Važnejši vplivi: padec tlaka v smeri roka pri enakem preseku cevi P 4 ' v L Padec tlaka zaradi trenja na dolžini cevi L.. aktor trenja (Fanning), označeno tudi C V literaturi so različni podatki za vrednosti aktorja trenja pogosto najdemo naslednjo zvezo: 4 = P v L Padec tlaka zaradi trenja na dolžini cevi L.. aktor trenja (Moody - arcy) Energetske izgube zaradi trenja tekočine ob steno cevi so odvisne od Re števila Za majhna Re števila laminarni tok 0 < Re < 300, (in prehodno območje) velja linearna zveza med Re številom in aktorjem trenja Laminarni tok 64 Re 64 v Mejna plast tekočine ob steni cevi je stacionarna, zato hrapavost cevi ne vpliva na aktor trenja 19
20 Laminarni turbulenti tok Padec tlaka zaradi trenja lahko izrazimo tudi z dinamično višino tekočine h : v L h = P/. g, pri čemer je P h h L v g arcy-weisbach-ova enačba v L V laminarnem toku hrapavost cevi (e) ne vpliva na padec tlaka, oz. na vrednost aktorja trenja, ker je mejna plast tekočine ob steni cevi je stacionarna. = (Re). Turbulentni tok: ni analitičnih rešitev za izračun padca tlaka zaradi zapletene hidrodinamske situacije. Rešitev so empirične enačbe, določene na osnovi eksperimentalnih podatkov. = (Re,,L,e,v,) Re 0. 5 Za popolnoma gladke cevi velja Blasius-ova enačba (1911): 3000 < Re <
21 Turbulenti tok Vpliv hrapavosti cevi na aktor trenja v turbulentnem toku. Hrapavosti cevi povzroča vrtinčast tok, ki poveča trenje tekočine med steno cevi in tekočino. Pri turbulentnem pretakanju tekočin po zaprti cevi na nastanek turbulence vpliva tudi hrapavost cevi. = (Re,e) Hrapavost e je deinirana v mm neravnih delov cevi in je navadno zelo majhna. Za različne materiale je hrapavost različna. Za določitev aktorja trenja je treba deinirati relativno hrapavost cevi: relativna hrapavost = hrapavost e premer cevi Enačbe za izračun aktorja trenja v turbulentnem toku so najpogosteje določene empirično na osnovi eksperimentov. Upoštevajo tudi hrapavost cevi. = (Re,,L,e,v,) Povprečne hrapavosti e komercialnih cevi Steel tube Wrought iron tube Copper tubing Glass tubing Polythene Flexible P.V.C. Rigid P.V.C. Cast iron tube Concrete tube Galvanised iron Wood stave mm mm mm mm mm mm mm mm.0000 mm (0.3-3) mm mm 1
22 Lewis Moody, 1944 Vpliv hrapavosti cevi na aktor trenja v turbulentnem toku. V turbulentnem področju so bile določene različne krivulje (na osnovi dimenzijske analize in z uporabo brezbimenzijskih števil), ki prikazujejo odvisnost aktorja trenja od Re števila in relativne hrapavosti (e/). Colebrookova enačba 64 Re
23 Moody, diagram 64 Re Re 0.5 3
24 Turbulenti tok Enačbe za izračun aktorja trenja v turbulentnem toku, ki so določene empirično na osnovi eksperimentov: Faktor trenja se najpogosteje določa iz diagramov, (Moody-jev diagram). V turbulentnem področju krivulje predstavljajo Colebrookovo enačbo. Za odčitek vrednosti aktorja trenja je treba izračunati Re število in poznati relativno hrapavost V laminarnem področj je = 64/Re Za popolnoma gladke cevi v turbulentnem področju velja Blasinusova zveza: = Re -0.5 Moddyjev diagram sta v logaritemskih koordinatah,treba je znati odčitati vrednosti 4
25 Primer: Izračun aktorja trenja: Po jekleni cevi premera 0.4 m in je dolga 10 m se pretaka voda s pretokom L/s. Temperatura vode je 10 C. oloči aktor trenja in padec tlaka zaradi trenja tekočine ob stene cevi! Snovne lastnosti vode določimo iz priročnika: = 1000 kg/m 3 =1.3 x 10-3 Pa.s Relativna hrapavost odčitamo iz tabele: e/ = /0.4 = Povprečne hrapavosti e komercialnih cevi Steel tube Wrought iron tube Copper tubing Glass tubing Polythene Flexible P.V.C. Rigid P.V.C. Cast iron tube Concrete tube Galvanised iron Wood stave mm mm mm mm mm mm mm mm.0000 mm (0.3-3) mm mm Povprečna hitrost toka: v = F v /A = x 4/ x 0.4 =.778 m/s Temp. Viscosity Speciic heat ensity ( C) (N s m - ) (kj kg -1 C -1 ) (kg m -3 ) x x x x x x x x A= x 0.4 /4 = m Re = x v x / = (1000 x.778 x 0.4) / =
26 Graična določitev aktorja trenja: P v L P P 1350 Pa =0.014 e/ = Re = h h L v g m 9.81 v 6
27 Primer: Izračunaj padec tlaka v cevi Po cevovodu s premerom 5 cm teče olivno olje spretokom 0.1m 3 /min. Izračunaj padec tlaka zaradi tekočinskega trenja, če je dolžina cevovoda 170 m in temperatura olivnega olja 0 C. 170 m A = x 0.05 /4 = x 10-3 m Iz priročnika odčitamo snovne lastnosti olivnega olja pri 0 C =910 kg/m 3 = Pa.s (N.s/m) Povprečna hitrost toka: v = F v /A = [0.1/60 (m3/s)] / m = 0.85 m/s Re = x v x / = (910 x 0.85 x 0.05) / = 460 Laminaren tok: = 64/Re = (Moddy-jev diagram- ne obsega tega območja) v L P P Pa
28 = 64/Re = (Moddy-jev diagram- ne obsega tega območja) 8
29 Primer: črpanje mleka Ocenite potrebno moč črpalke za črpanje mleka pri 0 C po 130 m dolgi horizontalni jekleni cevi premera 4 cm, če je povprečna hitrost toka v cevi.7 m/s.pri tem upoštevajte energetske izgube zaradi viskoznega trenja mleka ob steno cevi, energetske izgube na zaradi kolen in ventilov na cevovodu pa zanemari! L = 130m = 0.04m presek cevi: A = x 0.04 /4 = 1.6 x10-3 m Povprečna hitrost: v =.7 m/s volumski pretok F v = v x A =1.6x10-3 x.7 = m 3 /s Re = x v x / = (1030 x.7 x 0.04) /.1 x10-3 = x 10 4 Turbulentni tok Kinetična energija P k : Izgube zaradi viskoznega trenja: odčitamo hrapavost e= m e/ = /0.04 = Frikcijski aktor, aktor trenj, določimo iz Moodyjevega diagrama v Pa Steel tube Wrought iron tube Copper tubing Glass tubing Polythene Flexible P.V.C. Rigid P.V.C. Cast iron tube Concrete tube Galvanised iron Wood stave Iz priročnika poiščemo snovne lastnosti mleka: =.1 x 10-3 Pa.s = 1030 kg/m mm mm mm mm mm mm mm mm.0000 mm (0.3-3) mm mm
30 P v L Moč potrebna zaradi viskoznih izgub: P tr x F v = 3.05 x10 5 N/m x 3.4 x10-3 m 3 /s = 1038 J/s = W Moč potrošena zaradi kinetične energije: x 10 3 x3.4x10-3 = 1.76 w Celokupna moč: U + K = = W 5 Pa = 0.05 e/ = x
31 Viskozni vplivi: Važnejši vplivi: padec tlaka v smeri roka pri enakem preseku cevi aktor trenja h P v L h L v g arcy-weisbach-ova enačba L v Manjši vplivi: padec tlaka pri toku skozi ventile, razcepe, kolena in drugih sprememb preseka cevi. Če tok tekočine spremeni smer (kolena) ali je oviran zaradi vstavljenih ventilov v cevovode, se razcepi ali pa pride do zožitve cevi, pride do dodatnih energetskih izgub. Ta energija se porablja zaradi dodatnega vrtinčenja ali turbulence, oz. se sprošča v obliki toplote. Če želimo obdržat želeni pretok, moramo to energijo nadomestiti. Tudi ti, t.i. manjši viskozni vplivi na energetske izgube so sorazmerni dinamičnemu tlaku. P m 1 v High pressure R Low pressure 31
32 Viskozni vplivi: Manjši vplivi: padec tlaka pri toku skozi ventile, razcepe, kolena in drugih sprememb preseka cevi. Energetske izgube splošno: P / V : enota tlak Pa = N/m volumen: m 3 P / V = Nm = J Energetske izgube zaradi prej naštetih elementov lahko izrazimo z tlačno razliko, ali hodrostatsko višino. Pri tem smo uporabili prej izpeljane zveze za aktor trenja, in dodali aktor k. P m v k h m = P m /.g h m v k g P Faktor k je odvisen od tipa ventila, razcepa kolena in je za navedeno opremo določen eksperimentalno. Ker imamo v tehnološkem procesu več elementov, na primer več ventilov, kolen, razcepov, je za treba energetske izgube na posameznih elementih sešteti: v hm k g Celotne energetske izgube zaradi viskoznega trenja lahko izrazimo: L v h k g h v L 3 v g L
33 Faktor k za različne elemente v procesu: v hm k g Valves, ully open: k gate 0.13 globe 6.0 angle 3.0 Elbows: 90 standard 0.74 medium sweep 0.5 long radius 0.5 square 1.5 Tee, used as elbow 1.5 Tee, straight through 0.5 Entrance, large tank to pipe**: sharp 0.5 rounded
34 Entrance, large tank to pipe**: sharp 0.5 rounded 0.05 h m v k g a b c d Entrance low conditions and loss coeicient. (a) Re-entrant, k = 0.8, (b) sharp-edged, k = 0.5, (c) slightly rounded, k = 0., (d) well-rounded, k = Source: Munson et al (1998) p
35 v hm k g Energetske izgube nastopijo tudi zaradi zožitve ali razširitve cevi. Pri povečanju premera cevi so energetske izgube: P =. (v 1 - v ) / Pri čemer je v 1 je hitrost toka v nasprotni smeri spremembe premera cevi ( 1 ) in v hitrost toka v smeri spremembe premera cevi ( ) Pri trenutni zožitvi cevi pa: P= k. v / Faktor trenja k je odvisen od razmerja premerov ( / 1 ): Koeiicient k v primeru zožitve cevi v odvisnosti od razmerja premerov / k Loss coeicient or a sudden contraction. 35 Source: Munson et al (1998) p. 500
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