CITY AND GUILDS 9210 Level 6 Module - Unit 129 FLUID MECHANICS

Transcription

1 ITY AND GUILDS 90 Level 6 Mdule - Unt 9 FLUID MEHANIS OUTOME - TUTOIAL FLUID FLOW ALULATIONS Ths mdule has Leanng Outcmes. Ths s the secnd tutal utcme Outcme Pem lud lw calculatns The leane can:. Slve cmessble lud lw blems nvlvng: a seed weak essue waves b stagnatn essue c lud temeatue d lud densty.. Slve blems nvlvng sentc lw a eect gas n ducts vayng css-sectnal aea n tems Mach numbe and ncludng chked lw. 3. Descbe the matn a nmal shck n cnvegent-dvegent nzzles.. Detemne and aly lamna lw n es and n and between lat lates. 5. alculate the velcty dstbutn n lamna lw. 6. alculate the vlumetc lw ate n lamna lw. 7. Aly lamna lw t hyddynamc lubcatn. 8. Analyse lamna lw usng: a bunday laye they b dslacement and mmentum thcknesses c skn ctn cecent. 9. Slve blems usng the mmentum ntegal equatn. 0. alculate the dag n a lat late n lamna lw.. Descbe the acts aectng bunday laye tanstn.. Analyse tubulent bunday layes n tems : a we law b lgathmc velcty dstbutn c lamna sub-laye d skn ctn n a lat late. 3. alculate the dag n a lat late n tubulent lw.. Detemne and aly the eects suace ughness n lud lw. 5. Descbe bunday laye seaatn and the matn wakes. 6. Slve blems nvlvng steady lw n es : a Newtnan luds b nn-newtnan luds. 7. Analyse the elatnsh n steady lw between ctn act, eynlds numbe and elatve ughness. 8. Analyse smle e netwks usng teatve calculatns. 9. Aly Eule and Benull equatns t ncmessble nvscd lud lws. 0. Detemne and aly the steam unctn and velcty tental unctn n steady tw dmensnal lws.. Detemne and aly lws ncmessble luds esultng m smle cmbnatns : a unm steam b suce c snk d dublet e nt vte.. Detemne and aly nvscd lw aund a ccula cyde wth cculatn ncludng the calculatn a essue dstbutn b lt ce. Pe-equste Knwledge equement In de t study ths mdule yu shuld aleady have a gd knwledge lud mechancs. I nt yu shuld study the tutals at bee cmmencng ths mdule. We wll stat ths tutal by evsng vscsty and bunday layes. Ths was cveed n tutal utcme. D.J.Dunn

2 . VISOSITY In tutal utcme t was elaned that vscus luds have mlecules that tend t stck tgethe and t any suace wth whch t s n cntact (but nt always tue). Ths leads t the dea that luds n mtn must shea and that a ce s needed t make ths haen and ths aeas as lud ctn. Ths led t the dentn Newtnan Fluds shea stess dy DYNAMI VISOSITY = ate shea du The symbl s als cmmnly used dynamc vscsty. Fgue I the lud stcks t a suace (wets t) then the velcty the lud bulds u m ze whee t s stuck t the suace t a mamum and the mula deved the velcty was d u u y dl Ths s ltted n gue. The bunday laye thckness δ s dened as the value y whee the velcty eaches 99% the mansteam velcty u. UNITS VISOSITY Fgue The unts dynamc vscsty N s/m but the S.I. unt s the cent-pse cp = 0.00 N s/m. Als cmmnly used s the knematc vscsty dened as = µ/ The unts knematc vscsty ae m/s but the S.I. unt s the cent-stke (cst) cst = m/s = mm/s Nw lets study lamna and tubulent lw they. D.J.Dunn

3 . LAMINA FLOW THEOY The llwng wk nly ales t Newtnan luds. LAMINA FLOW A steam e s an magnay e wth n lw nmal t t, nly ag t. When the lw s lamna, the steames ae aallel and lw between tw aallel suaces we may cnsde the lw as made u aallel lamna layes. In a e these lamna layes ae cydcal and may be called steam tubes. In lamna lw, n mng ccus between adjacent layes and t ccus at lw aveage velctes. TUBULENT FLOW Fgue 3 The sheang cess causes enegy lss and heatng the lud. Ths nceases wth mean velcty. When a cetan ctcal velcty s eceeded, the steames beak u and mng the lud ccus. The dagam llustates eynlds clued bbn eement. lued dye s njected nt a hzntal lw. When the lw s lamna the dye asses ag wthut mng wth the wate. When the seed the lw s nceased tubulence sets n and the dye mes wth the suundng wate. One elanatn ths tanstn s that t s necessay t change the essue lss nt the ms enegy such as angula knetc enegy as ndcated by small eddes n the lw. LAMINA AND TUBULENT BOUNDAY LAYES It was elaned that a bunday laye s the laye n whch the velcty gws m ze at the wall (n sl suace) t 99% the mamum and the thckness the laye s dented. When the lw wthn the bunday laye becmes tubulent, the shae the bunday layes waves and when dagams ae dawn tubulent bunday layes, the mean shae s usually shwn. mang a lamna and tubulent bunday laye eveals that the tubulent laye s thnne than the lamna laye. Fgue D.J.Dunn 3

4 ITIAL VELOITY - EYNOLDS NUMBE Q When a lud lws n a e at a vlumetc lw ate Q m3/s the aveage velcty s dened u m A A s the css sectnal aea. u md u md The eynlds numbe s dened as e I yu check the unts e yu wll see that thee ae nne and that t s a dmensnless numbe. Ths was cveed n tutal 3 utcme. eynlds dscveed that t was ssble t edct the velcty lw ate at whch the tanstn m lamna t tubulent lw ccued any Newtnan lud n any e. He als dscveed that the ctcal velcty at whch t changed back agan was deent. He und that when the lw was gadually nceased, the change m lamna t tubulent always ccued at a eynlds numbe 500 and when the lw was gadually educed t changed back agan at a eynlds numbe 000. Nmally, 000 s taken as the ctcal value. WOKED EXAMPLE Ol densty 860 kg/m 3 has a knematc vscsty 0 cst. alculate the ctcal velcty when t lws n a e 50 mm be damete. SOLUTION u e m u md ν eν D m/s D.J.Dunn

5 DEIVATION OF POISEUILLE'S EQUATION LAMINA FLOW Pseulle dd the gnal devatn that llws, It elates essue lss n a e t the velcty and vscsty LAMINA FLOW. Hs equatn s the bass measuement vscsty hence hs name has been used the unt vscsty. nsde a e wth lamna lw n t. nsde a steam tube length L at adus and thckness d. y s the dstance m the e wall. du du y dy d dy d Fgue 5 The shea stess n the utsde the steam tube s. The ce (F s ) actng m ght t let s due t the shea stess and s und by multlyng by the suace aea. Fs = L du du F a Newtnan lud. Substtutng we get F s dy d The essue deence between the let end and the ght end the sectn s. The ce due t ths (F ) s ccula aea adus. F = du - L d du Equatng ces we have - L du d d L In de t btan the velcty the steame at any adus we must ntegate between the lmts u = 0 when = and u = u when =. u du - d L 0 u L L Ths s the equatn a Paabla s the equatn s ltted t shw the bunday laye, t s seen t etend m ze at the edge t a mamum at the mddle. F mamum velcty ut = 0 and we get u L The aveage heght a aabla s hal the mamum value s the aveage velcty s u m Fgue 6 8 L Oten we wsh t calculate the essue d n tems damete D. Substtute =D/ and eaange. 3Lum D The vlume lw ate s aveage velcty css sectnal aea. D Q 8L 8L 8L Ths s ten changed t gve the essue d as a ctn head. 3Lum The ctn head a length L s und m h =/g h gd Ths s Pseulle's equatn that ales nly t lamna lw. D.J.Dunn 5

6 WOKED EXAMPLE A callay tube s 30 mm g and mm be. The head equed t duce a lw ate 8 mm3/s s 30 mm. The lud densty s 800 kg/m3. alculate the dynamc and knematc vscsty the l. SOLUTION eaangng Pseulle's equatn we get h gd 3Lu m d A mm Q 8 u m 0.8 mm/s A N s/m.cp m / s 30.cSt 800 WOKED EXAMPLE 3 Ol lws n a e 00 mm be wth a eynlds numbe 50. The dynamc vscsty s 0.08 Ns/m. The densty s 900 kg/m3. Detemne the essue d e mete length, the aveage velcty and the adus at whch t ccus. SOLUTION e=u m D/. Hence u m = e / D u m = ( )/(900 0.) = 0.05 m/s = 3µL u m /D = /0. =.88 Pascals. u = {/Lµ}( - ) whch s made equal t the aveage velcty 0.05 m/s 0.05 = (.88/ 0.08)( ) = m 35.3 mm. D.J.Dunn 6

7 FLOW BETWEEN FLAT PLATES nsde a small element lud mvng at velcty u wth a length d and heght dy at dstance y abve a lat suace. The shea stess actng n the element nceases by d n the y dectn and the essue deceases by d n the dectn. d d u It was shwn eale that d dy It s assumed that d/d des nt vay wth y s t may be egaded as a ed value n the llwng wk. d du Integatng nce - y A d dy Fg. 7 y d Integatng agan - u Ay B...( A) d A and B ae cnstants ntegatn. The slutn the equatn nw deends un the bunday cndtns that wll yeld A and B. WOKED EXAMPLE Deve the equatn kng velcty u and heght y at a gven nt n the dectn when the lw s lamna between tw statnay lat aallel lates dstance h aat. G n t deve the vlume lw ate and mean velcty. SOLUTION When a lud tuches a suace, t stcks t t and mves wth t. The velcty at the lat lates s the same as the lates and n ths case s ze. The bunday cndtns ae hence u = 0 when y = 0 Substtutng nt equatn.6a yelds that B = 0 u=0 when y=h Substtutng nt equatn.6a yelds that A = (d/d)h/ Puttng ths nt equatn.6a yelds u = (d/d)(/){y - hy} (The student shuld d the algeba ths). The esult s a aablc dstbutn smla that gven by Pseulle's equatn eale nly ths tme t s between tw lat aallel suaces. D.J.Dunn 7

8 FLOW ATE T nd the lw ate we cnsde lw thugh a small ectangula slt wdth B and heght dy at heght y. Fgue 8 The lw thugh the ed slt nmal t the age s dq = u Bdy =(d/d)(/){y - hy} Bdy Integatng between y = 0 and y = h t nd Q yelds Q = -B(d/d)(h3/) The mean velcty s um = Q/Aea = Q/Bh hence um = -(d/d)(h/) (The student shuld d the algeba) OTATING ONENTI YLINDES nsde eamle a shat tatng cncentcally nsde anthe cyde lled wth l. Thee s n veall lw ate s equatn (A) des nt aly. Due t the stckness the lud, the lqud stcks t bth suaces. On the shat suace the velcty s u = and at the ute suace t s ze. I the ga s small, t may be assumed that the change n the velcty acss the ga changes m u t ze ealy wth adus. Fgue 9 = µ du/dy But snce the change s ea du/dy = u/( - ) = /( - ) s = µ /( - ) Shea ce n cyde F shea stess suace aea 3 h h F h Tque F T F T In the case a tatnal vscmete we eaange s that 3 h In ealty, t s unlkely that the velcty vaes ealy wth adus and the bttm the cyde wuld have an aect n the tque. I ths was a beang (e.g. shat nsde a bush) t wuld be an eamle hydstatc lubcatn. D.J.Dunn 8

9 FALLING SPHEES Ths they may be aled t atcle seaatn n tanks and t a alg shee vscmete. When a shee alls, t ntally acceleates unde the actn gavty. The esstance t mtn s due t the sheang the lqud assng aund t. At sme nt, the esstance balances the ce gavty and the shee alls at a cnstant velcty. Ths s the temnal velcty. F a bdy mmesed n a lqud, the buyant weght s W and ths s equal t the vscus esstance when the temnal velcty s eached. 3 d g s = W = vlume densty deence gavty 6 s = densty the shee mateal = densty lud d = shee damete Fgue 0 The vscus esstance s much hade t deve m st ncles and ths wll nt be attemted hee. In geneal, we use the cncet DAG and dene the DAG OEFFIIENT as esstance ce D Dynamc essue jected Aea u d The dynamc essue a lw steams The jected aea a shees eseach shws the llwng elatnsh between D and e a shee. D 8 u d Fgue F e <0. the lw s called Stkes lw and Stkes shwed that = 3du hence D =/ ud = / e F 0. < e < 500 the lw s called Allen lw and D =8.5 e -0.6 F 500 < e < 0 5 D s cnstant D = 0. An emcal mula that cves the ange 0. < e < 0 5 s as llws. D 6 e e 0. F a alg shee vscmete, Stkes lw ales. Equatng the dag ce and the buyant weght we get 3du = (d 3 /6)( s - ) g = gd ( s - )/8u a alg shee vscmete The temnal velcty Stkes lw s u = d g( s - )8 Ths mula assumes a lud nnte wdth but n a alg shee vscmete, the lqud s squeezed between the shee and the tube walls and addtnal vscus esstance s duced. The Faen cectn act F s used t cect the esult. D.J.Dunn 9

10 THUST BEAINGS nsde a und lat dsc adus tatng at angula velcty ad/s n t a lat suace and seaated m t by an l lm thckness t. Ths s an eamle HYDOSTATI LUBIATION. Fgue Assume the velcty gadent s ea n whch case du/dy = u/t = /t at any adus. du Theshea stessn the ng s dy t Theshea ce s df d t 3 The tque s dt df d t The ttal tque s und by ntegatng wth esect t. T 0 3 d t t In tems damete D ths s T D 3t Thee ae many vaatns n ths theme that yu shuld be eaed t handle. D.J.Dunn 0

11 MOE ON FLOW THOUGH PIPES nsde an elementay thn cydcal laye that makes an element lw wthn a e. The length s, the nsde adus s and the adal thckness s d. The essue deence between the ends s and the shea stess n the suace nceases by d m the nne t the ute suace. The velcty at any nt s u and the dynamc vscsty s. Fgue 3 The essue ce actng n the dectn lw s The shea ce sng s {( + d) - } {( + )()( + d) - } Equatng, smlyng and gnng the duct tw small quanttes we have the llwng esult. d du Newtnan luds. d dy I y s du d u d d du d u d d du d d Usng atal deentatn t deentate d du d d hence d Integatng we get du d measued m thensde the ethen d u d A du A d...(a) whee A s a cnstant ntegatn. Integatng agan we get u A B...( B) whee B s anthe cnstant ntegatn. du d -y and dy yelds the esult du - d s d du d u d d Equatns (A) and (B) may be used t deve Pseulle's equatn t may be used t slve lw thugh an annula assage. D.J.Dunn

12 D.J.Dunn PIPE At the mddle =0 s m equatn (A) t llws that A = 0 At the wall, u=0 and =. Puttng ths nt equatn B yelds agan. s equatn and ths spseulle' 0 whee A 0 u B B A ANNULUS Fgue A A B A B A B A u 0 0 subtact D m...(d) 0 )...( 0. and at 0 ae u cndtns The bunday 0 B nt equatn ut Ths s 0 be btaned m. may be substtuted back nt equatn D.The same esult wll Ths A B B

13 D.J.Dunn 3 u - u u F gven values the velcty dstbutn s smla t ths. Fgue 5

14 SELF ASSESSMENT EXEISE N.. Ol lws n a e 80 mm be damete wth a mean velcty 0. m/s. The densty s 890 kg/m3 and the vscsty s Ns/m. Shw that the lw s lamna and hence deduce the essue lss e mete length. (50 Pa e mete).. Ol lws n a e 00 mm be damete wth a eynlds Numbe 500. The densty s 800 kg/m3. alculate the velcty a steame at a adus 0 mm. The vscsty µ = 0.08 Ns/m. (0.36 m/s) 3. A lqud dynamc vscsty Ns/m lws thugh a callay damete 3.0 mm unde a essue gadent 800 N/m3. Evaluate the vlumetc lw ate, the mean velcty, the cente e velcty and the adal stn at whch the velcty s equal t the mean velcty. (uav = 0.0 m/s, uma = 0.0 m/s =.06 mm). Smla t E.. Eam Q6 998 a. Elan the tem Stkes lw and temnal velcty. b. Shw that a shecal atcle wth Stkes lw has a temnal velcty gven by u = d g( s - )/8 G n t shw that D =/ e c. F shecal atcles, a useul emcal mula elatng the dag cecent and the eynld s numbe s 6 D 0. e Gven = 000 kg/m 3, = cp and s = 630 kg/m 3 detemne the mamum sze shecal atcles that wll be lted uwads by a vetcal steam wate mvng at m/s. d. I the wate velcty s educed t 0.5 m/s, shw that atcles wth a damete less than 5.95 mm wll all dwnwads. e D.J.Dunn

15 5. Smla t E.. Eam Q5 998 A smle lud cug cnssts tw aallel und dscs Damete D seaated by a ga h. One dsc s cnnected t the nut shat and tates at ad/s. The the dsc s cnnected t the utut shat and tates at ad/s. The dscs ae seaated by l dynamc vscsty and t may be assumed that the velcty gadent s ea at all ad. D Shw that the Tque at the nut shat s gven by T 3h The dscs ae 300 mm damete and the ga s. mm. The nut shat tates at 900 ev/mn and tansmts 500W we. alculate the utut seed, tque and we. (77 ev/mn, 5.3 Nm and W) Shw by alcatn ma/mn they that the utut seed s hal the nut seed when mamum utut we s btaned. 6. Shw that ully develed lamna lw a lud vscsty between hzntal aallel lates a dstance h aat, the mean velcty um s elated t the essue gadent d/d by Fgue 6 shws a langed e jnt ntenal damete d cntanng vscus lud vscsty at gauge essue. The lange has an ute damete d and s meectly tghtened s that thee s a naw ga thckness h. Obtan an eessn the leakage ate the lud thugh the lange. Fgue 6 Nte that ths s a adal lw blem and B n the ntes becmes and d/d becmes -d/d. An ntegatn between nne and ute ad wll be equed t gve lw ate Q n tems essue d. The answe s D.J.Dunn 5

16 3. TUBULENT FLOW FITION OEFFIIENT The ctn cecent s a cnvenent dea that can be used t calculate the essue d n a e. It s dened as llws. Wall Shea Stess Dynamc Pessue DYNAMI PESSUE nsde a lud lwng wth mean velcty u m. I the knetc enegy the lud s cnveted nt lw lud enegy, the essue wuld ncease. The essue se due t ths cnvesn s called the dynamc essue. KE = ½ mu m Flw Enegy = Q Q s the vlume lw ate and = m/q Equatng ½ mu m = Q = mu /Q = ½ u m WALL SHEA STESS The wall shea stess s the shea stess n the laye lud net t the wall the e. The shea stess n the laye net t the wall s The shea ce esstng lw s F s LD Fgue 7 du dy D The esultng essue d duces a ce F D Equatng ces gves L FITION OEFFIIENT LAMINA FLOW wall Wall Shea Stess Dynamc Pessue D Lu 3Lu m Fm Pseulle s equatn Hence D m D Lu m 3Lu 6 D u md 6 e D.J.Dunn 6

17 DAY FOMULA Ths mula s manly used calculatng the essue lss n a e due t tubulent lw but t can be used lamna lw als. Tubulent lw n es ccus when the eynlds Numbe eceeds 500 but ths s nt a clea nt s 3000 s used t be sue. In de t calculate the ctnal lsses we use the cncet ctn cecent symbl. Ths was dened as llws. Wall Shea Stess Dynamc Pessue D Lu eaangng equatn t make the subject Lu m D Ths s ten eessed as a ctn head h Lu m h g gd Ths s the Dacy mula. In the case lamna lw, Dacy's and Pseulle's equatns must gve the same esult s equatng them gves Lu m 3Lu m gd gd Ths s the same esult as bee lamna lw. 6 u D Tubulent lw may be saely assumed n es when the eynlds Numbe eceeds In de t calculate the ctnal lsses we use the cncet ctn cecent symbl. Nte that n lde tetbks was wtten as but nw the symbl eesents. FLUID ESISTANE Flud esstance s an altenatve aach t slvng blems nvlvng lsses. The abve equatns may be eessed n tems lw ate Q by substtutng u = Q/A m 6 e m h Lu gd LQ m Substtutng A =D / we get the llwng. gda h 3 Q 5 s the lud esstance estctn. g LQ D 3 L g D 5 I we want essue lss nstead head lss the equatns ae as llws. 3 gh Q s the lud esstance estctn. 5 D LQ 3 L 5 D It shuld be nted that cntans the ctn cecent and ths s a vaable wth velcty and suace ughness s shuld be used wth cae. D.J.Dunn 7

18 MOODY DIAGAM AND ELATIVE SUFAE OUGHNESS In geneal the ctn head s sme unctn u m such that h = u m n. lealy lamna lw, n = but tubulent lw n s between and and ts ecse value deends un the ughness the e suace. Suace ughness mtes tubulence and the eect s shwn n the llwng wk. elatve suace ughness s dened as = k/d whee k s the mean suace ughness and D the be damete. An Amecan Engnee called Mdy cnducted ehaustve eements and came u wth the Mdy hat. The chat s a lt vetcally aganst e hzntally vaus values. In de t use ths chat yu must knw tw the thee c-dnates n de t ck ut the nt n the chat and hence ck ut the unknwn thd cdnate. F smth es, (the bttm cuve n the dagam), vaus mulae have been deved such as thse by Blasus and Lee. BLASIUS = e 0.5 LEE = e The Mdy dagam shws that the ctn cecent educes wth eynlds numbe but at a cetan nt, t becmes cnstant. When ths nt s eached, the lw s sad t be ully develed tubulent lw. Ths nt ccus at lwe eynlds numbes ugh es. A mula that gves an amate answe any suace ughness s that gven by Haaland lg 0 e D.J.Dunn 8

19 D.J.Dunn Fgue 8 HAT 9

20 WOKED EXAMPLE Detemne the ctn cecent a e 00 mm be wth a mean suace ughness 0.06 mm when a lud lws thugh t wth a eynlds numbe SOLUTION The mean suace ughness = k/d = 0.06/00 = Lcate the e = k/d = Tace the e untl t meets the vetcal e at e = ead the value hzntally n the let. Answe = heck usng the mula m Haaland. 3.6 lg 3.6 lg 3.6 lg e WOKED EXAMPLE 5 Ol lws n a e 80 mm be wth a mean velcty m/s. The mean suace ughness s 0.0 mm and the length s 60 m. The dynamc vscsty s N s/m and the densty s 900 kg/m 3. Detemne the essue lss. SOLUTION e = ud/ = ( )/0.005 = = k/d = 0.0/80 = Fm the chat = h = Lu/dg = ( )/( ) =.7 m = gh = =.3 kpa. (c) Auth D.J.Dunn 0

21 SELF ASSESSMENT EXEISE N.. A e s 5 km g and 80 mm be damete. The mean suace ughness s 0.03 mm. It caes l densty 85 kg/m3 at a ate 0 kg/s. The dynamc vscsty s 0.05 N s/m. Detemne the ctn cecent usng the Mdy hat and calculate the ctn head. (Ans m.). Wate lws n a e at 0.05 m3/s. The e s 50 mm be damete. The essue d s 3 0 Pa e mete length. The densty s 000 kg/m3 and the dynamc vscsty s 0.00 N s/m. Detemne. the wall shea stess (67.75 Pa). the dynamc essue (980 Pa).. the ctn cecent ( ) v. the mean suace ughness ( mm) 3. Elan bely what s meant by ully develed lamna lw. The velcty u at any adus n ully develed lamna lw thugh a staght hzntal e ntenal adus s gven by d/d s the essue gadent n the dectn lw and µ s the dynamc vscsty. Shw that the essue d ve a length L s gven by the llwng mula. The wall skn ctn cecent s dened as Shw that = 6/e whee e = umd/µ and s the densty, um s the mean velcty and s the wall shea stess.. Ol wth vscsty 0- Ns/m and densty 850 kg/m3 s umed ag a staght hzntal e wth a lw ate 5 dm3/s. The statc essue deence between tw tang nts 0 m aat s 80 N/m. Assumng lamna lw detemne the llwng.. The e damete.. The eynlds numbe. mment n the valdty the assumtn that the lw s lamna (c) Auth D.J.Dunn

22 . NON-NEWTONIAN FLUIDS The st at ths sectn s a eeat the wk n tutal utcme s yu wsh g n t the sectn n lastc lw. A Newtnan lud as dscussed s a n ths tutal s a lud that beys the law A Nn Newtnan lud s geneally descbed by the nn-ea law y s knwn as the yeld shea stess and s the ate shea stan. nsde the ncle ms ths equatn shwn n the net dagam. Fgue 9 Gah A shws an deal lud that has n vscsty and hence has n shea stess at any nt. Ths s ten used n theetcal mdels lud lw. Gah B shws a Newtnan Flud. Ths s the tye lud wth whch ths bk s mstly cncened, luds such as wate and l. The gah s hence a staght e and the gadent s the vscsty. Thee s a ange the lqud sem-lqud mateals that d nt bey ths law and duce stange lw chaactestcs. Such mateals nclude vaus dstus, ants, cements and s n. Many these ae n act sld atcles susended n a lqud wth vaus cncentatns. Gah shws the elatnsh a Dlatent lud. The gadent and hence vscsty nceases wth and such luds ae als called shea-thckenng. Ths henmenn ccus wth sme slutns suga and staches. Gah D shws the elatnsh a Pseud-lastc. The gadent and hence vscsty educes wth and they ae called shea-thnnng. Mst dstus ae lke ths as well as clay and lqud cement. Othe luds behave lke a lastc and eque a mnmum stess y bee t sheas. Ths s lastc behavu but unlke lastcs, thee may be n elastcty t sheang. Gah E shws the elatnsh a Bngham lastc. Ths s the secal case whee the behavu s the same as a Newtnan lud ecet the estence the yeld stess. Fdstus cntanng hgh level ats amate t ths mdel (butte, magane, chclate and Maynnase). Gah F shws the elatnsh a lastc lud that ehbts shea thckenng chaactestcs. Gah G shws the elatnsh a assn lud. Ths s a lastc lud that ehbts shea-thnnng chaactestcs. Ths mdel was develed luds cntanng d lke slds and s ten aled t mlten chclate and bld. (c) Auth D.J.Dunn

23 MATHEMATIAL MODELS The gahs that elate shea stess and ate shea stan ae based n mdels equatns. Mst ae mathematcal equatns ceated t eesent emcal data. Hschel and Bulkeley develed the we law nn-newtnan equatns. Ths s as llws. n y K K s called the cnsstency cecent and n s a we. In the case a Newtnan lud n = and y = 0 and K = (the dynamc vscsty) F a Bngham lastc, n = and K s als called the lastc vscsty. The elatnsh educes t y F a dlatent lud, y = 0 and n> F a seud-lastc, y = 0 and n< The mdel bth s The Hechel-Bulkeley mdel s Ths may be develed as llws. n K K dvdng by n y K K y K a y y n n y K smetmes wtten as n n F a Bngham lastcn s The at s called the aaent vscsty a y K F a Flud wth n yeld shea value 0 y y s n a whee s called K a n the lastc vscsty. The assn lud mdel s qute deent n m m the thes and s as llws. y K (c) Auth D.J.Dunn 3

24 (c) Auth D.J.Dunn THE FLOW OF A PLASTI FLUID Nte that luds wth a shea yeld stess wll lw n a e as a lug. Wthn a cetan adus, the shea stess wll be nsucent t duce sheang s nsde that adus the lud lws as a sld lug. Fgue 0 shws a tycal stuatn a Bngham Plastc. Fgue 0 MINIMUM PESSUE The shea stess actng n the suace the lug s the yeld value. Let the lug be damete d. The essue ce actng n the lug s d / The shea ce actng n the suace the lug s y d L Equatng we nd d / = y d L d = y L/ = y L/d The mnmum essue equed t duce lw must ccu when d s lagest and equal t the be the e. (mnmum) = y L/D The damete the lug at any geate essue must be gven by d = y L/ F a Bngham Plastc, the bunday laye between the lug and the wall must be lamna and the velcty must be elated t adus by the mula deved eale. L 6 L d D u FLOW ATE The lw ate shuld be calculated n tw stages. The lug mves at a cnstant velcty s the lw ate the lug s smly Q = u css sectnal aea = u d / The lw wthn the bunday laye s und n the usual way as llws. nsde an elementay ng adus and wdth d. L Q L L Q d L Q d L d u dq 3 The mean velcty as always s dened as u m = Q/ss sectnal aea.

25 WOKED EXAMPLE 5 The Hechel-Bulkeley mdel a nn-newtnan lud s as llws. n y K. Deve an equatn the mnmum essue equed d e mete length n a staght hzntal e that wll duce lw. Gven that the essue d e mete length n the e s 60 Pa/m and the yeld shea stess s 0. Pa, calculate the adus the slug sldng thugh the mddle. SOLUTION Fgue The essue deence actng n the css sectnal aea must duce sucent ce t vecme the shea stess actng n the suace aea the cydcal slug. F the slug t mve, the shea stess must be at least equal t the yeld value y. Balancng the ces gves the llwng. = y L /L = y / 60 = 0./ = 0./60 = m 6.6 mm (c) Auth D.J.Dunn 5

26 WOKED EXAMPLE 6 A Bngham lastc lws n a e and t s bseved that the cental lug s 30 mm damete when the essue d s 00 Pa/m. alculate the yeld shea stess. Gven that at a lage adus the ate shea stan s 0 s - and the cnsstency cecent s 0.6 Pa s, calculate the shea stess. SOLUTION F a Bngham lastc, the same they as n the last eamle ales. /L = y / 00 = y /0.05 y = / = 0.75 Pa A mathematcal mdel a Bngham lastc s K y = =.75 Pa (c) Auth D.J.Dunn 6

27 SELF ASSESSMENT EXEISE N. 3. eseach has shwn that tmat ketchu has the llwng vscus etes at 5. nsstency cecent K = 8.7 Pa s n Pwe n = 0.7 Shea yeld stess = 3 Pa alculate the aaent vscsty when the ate shea s, 0, 00 and 000 s - and cnclude n the eect the shea ate n the aaent vscsty. Answes = a = 50.7 = 0 a = 6.68 = 00 a = = 000 a = A Bngham lastc lud has a vscsty 0.05 N s/m and yeld stess 0.6 N/m. It lws n a tube 5 mm be damete and 3 m g. () Evaluate the mnmum essue d equed t duce lw. (80 N/m ) The actual essue d s twce the mnmum value. Sketch the velcty le and calculate the llwng. () The adus the sld ce. (3.75 mm) () The velcty the ce. (67.5 mm/s) (v) The vlumetc lw ate. (7.6 cm 3 /s) du 3. A nn-newtnan lud s mdelled by the equatn K whee n = 0.8 and d K = 0.05 N s 0.8 /m. It lws thugh a tube 6 mm be damete unde the nluence a essue d 600 N/m e mete length. Obtan an eessn the velcty le and evaluate the llwng. () The cente e velcty. (0.953 m/s) () The mean velcty. (0.5 m/s) n (c) Auth D.J.Dunn 7