Tribology Prof. Dr.Harish Hirani Department of Mechanical Engineering Indian Institute of Technology, Delhi. Lecture No. # 20 Reynolds Equation

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1 Trbology ro. Dr.Hars Hran Deartment o Mecancal Engneerng Indan Insttute o Tecnology, Del Lecture No. # 0 Reynolds Equaton Welcome to twentet lecture o vdeo course on trbology. Today s toc s te Reynolds equaton. In my revous lecture, we derved ts equaton; assumng ew concets or assumtons based on tose assumtons, equaton was derved, t s not generaled one, but restrcted to systems, wc ull te assumtons. Today, we are gong to dscuss ts equaton or some varatons o ts equaton. (Reer Slde Tme: 00:59) Reynolds Equaton So, rst ts slde sows Reynolds equaton, wc was derved n revous lecture, let and sde sows ressure term. We need to derve ressure dstrbuton over te suraces. Well, rgt and sde sows te source terms, tere are tree source terms, rst and last source term, were te dervaton o lm tckness s comng, tat s known as te wedge acton. Basc assumton s tat two suraces are nclned at some lne, t may be one degree, may be two degrees, or may be some degree. Some nclnaton s requred to develo te ressure tat s te wedge term. We can smly also te wedge term as dervaton o wt resect to. To nd or you say to dene te

2 wedge acton drecton smlarly, we can wrte wedge acton n drecton. In ts gure, we are sowng wedge acton n drecton. I I assume erendcular to ts tat wll be te drecton and varaton o lm tckness n drecton. Because o ts nclnaton tere s a ossblty o ressure generaton. You say tat or ydrodynamc acton wedge acton s essental. We drve te second term also wc s known as a squeee term. Bascally deends on a squeeng acton o lates o relatve movement along te lm tckness. I I am assumng te trbal base, tey move erendcular to eac oter. And ts gure llustrates tat yellow late as a varable vscosty n bot te drectons. Smlarly, red block, red late as a relatve velocty n between tere s a vscous lud. I yellow late s movng n ts y drecton, s gong to generate some ressure and t can be a ressure term. And trd source s agan stretcng term. But beore tat we ere, te V mnus V s a relatve velocty can be eressed also te varaton o wt resect to tme. As velocty canges, movement o ts late wll ncrease or decrease te lm tckness. And as velocty s a constant, tey say tme term comes nto te cture. Tat s wy we say, t can be eressed as artal varaton o wt resect to tme. As I mentoned trd term s a stretcng term, we are assumng lates are not sold lates. Tey stretc and vary te velocty along drecton and drecton tat s wy we are reresentng te stretcng term as varaton o U. Ten you sow velocty along te drecton wt resect to drecton or varaton n W wt resect to tat s reresented over ere. (Reer Slde Tme: 05:6) U W, η η ( U U ) ( W W ) ( V V ) ( U U ) ( U U ) ( U U ) ( W W ) ( W W ) ( W W )

3 To elaborate ts stretcng term, we ave one more slde to elan n te cases lke stretcng acton. I am tryng to sow wt eamle o one late, say tere s a velocty along te drecton, but, ts velocty s not constant. I I take derent sectons, say late as a secton, t as a velocty U, late as a velocty U, late at te oston as a velocty U. Smlarly, late at 8 as a velocty U 8. Wat s te meanng o ts? Ts late s not rgd watever te velocty over ts surace s not te same velocty o te slgtly aead o tat, t may be ncreasng or may be decreasng. As a stretcng term, tycal eamle s a rubber materal, wc s stretced. And we nd slgt varaton n velocty along te surace and tat s also a source o ressure generaton. So, ts wedge drecton n most o cases s neglgble. I you want to reresent n Reynolds equaton, wc was elaned n revous slde. We can wrte ts term or artly derentate ts term, rst derentaton wt resect to, derentaton o wt resect to and ten n second case derentaton o velocty wt resect to. So, ts term can be reresented o ts rst term can be reresented I as a relatve velocty dvde by two. And artal varaton o wt resect to lus ere s not derentated. So, by and derentaton not gve mnus U wt resect to. So, one term s converted nto two terms and wat s a stretcng term s te second term, rst term wll be wedge acton. So, ts term as a two comonents, one s a wedge term and second one s a stretcng term. Smlarly, n drecton ts term can be eressed n two terms. In rst case derentaton s only wt resect to derentaton o wt resect to and n second term derentaton o velocty wt resect to. I tnk tere s some mstake, ts wll be sad derentaton, wt resect to. So, ts wll be your to terms, varaton o velocty wt resect to and varaton o wt resect to.

4 (Reer Slde Tme: 08:8) Reynolds Equaton Now, comng back to te Reynolds equaton, we want to nd out te varous varatons, we derve ts equaton wt a number o assumtons. We say tat ntal term s a neglgble, neglgble o ressure gradent along te y drecton tat means, tere s no varaton along te y drecton. Watever we coose ressure wll reman same, at any y locaton, ten Newtonan lud we took as te eamle o Newtonan lud, we use relaton tan constant vscosty value, no sl neglectng angle o nclnaton or coordnate system and ncomressble low. Wen we are sayng tat tere s a constant value o te vscosty, wle n ts equaton s stll ts vscosty term s n a derental orm. As tese terms are te constant, t can be taken out easly wtout derentaton, and ts equaton can be rearranged. So, ts s wat we are sayng tat we take out ts η rearranged nto te rgt and sde, ts equaton wll be slgtly moded. We say now nstead o al ts term as te 6η, η dvde by two, t wll turn out to be 6η. Smlarly, n ts case tere s no multlcaton, but, ere ater takng ts n te rgt and sde s turnng out to be η nto d by dt. Here s agan only al multly wt η, t wll turn out to be 6η and term along te drecton, wc reresent te wedge acton as well as te stretcng acton. Now, to smly ts equaton, we assume veloctes along te drecton n oter word, tere s no relatve velocty n drecton. So, ts term can be neglected n ts stuaton.

5 Because, tere s no velocty allowed tere or tere s no relatve velocty along te drecton. In tat stuaton, ts equaton s smled, one term as been elmnated tat gves te smled eresson. Te ressure term on te let and sde and ts term eress te wedge acton lus stretcng acton and ts s squeee acton. Now, we ave a tree source terms over ere, one s a squeee acton term, one s wedge acton n drecton and stretcng acton n drecton. (Reer Slde Tme: :) We can urter smly ts equaton, by assumng bot te suraces are rgd suraces. Tere s no stretcng acton, we are not talkng about te rubber materal, we are talkng about more lke metals or ceramcs wc cannot be stretced. So, easly I know stress stretcng acton s neglgble or our calculaton. So, wt ts assumton te bot rgd suraces, ts term can be varaton o relatve velocty wt resect to can be neglected. And as we eressed n revous slde, revous to revous slde tat ts term can be eressed n two terms, one s stretc term and anoter wedge term. ( ) t U U 6η η ( ) t U U 6 η t U 6 η

6 Stretc term s neglgble, because o te rgd surace. So, ten only we wll be gettng te wedge term, tat s gven by denton and ts s one o te common Reynold equaton number ourt s smlcaton can be made. So, we are talkng about te relatve velocty. Agan assume one o te surace s nclned or artcularly, we say only nclned surace s s sldng at some velocty relatve to statonary surace. Wc s a stragt surace does not move and U can be neglected, U s 0. Wt tat knd o tng, we say tat only nclned surace sldes, equaton wll be moded U wll be 0 and n tat stuaton t wll be only U. As tere s only one velocty tere s no ont to say t s a U, U, U, we can say drectly t s a U (velocty o sldng surace). Wt tat equaton wll be moded n ts orm, tere s a ressure term n drecton, ressure term n drecton constant vscosty and ts s a relatve velocty o nclned surace n drecton. Ts s te wedge term n drecton and ts s te squeee term and ts s one o te most commonly used Reynolds equaton. In lterature, n number o books Reynolds equaton starts wt ts knd o orm. Were te number o assumtons are tere. In understand ts equaton, we ave made a number o assumtons. We assume tat s neglgble n a sort term, ts may not be necessary n artcularly g velocty terms. In a sort term wll not be neglgble, ten we assume te lm tckness drecton o ressure gradent s neglgble. Tat aens only wen tere s a very small lm tckness nntely small tckness. And we assure Newtonan lud and nterestng tng most o te lquds are non-newtonan lquds, tat tey are not Newtonan lquds. Ten we assume te constant value o te vscosty wc we know very well Vscosty wll not reman constant. It s a uncton I say t s a strong uncton o temerature ten tere s no sl on te lqud and sold surace. Wat do you do wt te sold surace? Is te relatve velocty n te same sldng, seed s marted to lqud wtout any loss. Wc most o te cases, most o te lqud cases s true, but, s not true or te gases, were te molecular dstance s muc and tere s earnest ossblty o te sl. Ten we assume te neglectng angle o nclnaton or te coordnate system n te angle o nclnaton s te very small ten tat s true, but, te angle o nclnaton s more ten t may not be true. We assume te ncomressble low, even n lqud cases many lquds can be comressed to ercent or ercent. So, assumng ncomressble low, rovdes may be 95 to 90 ercent

7 98 ercent good results, but not undred ercent results. I you requre really 00 accurate results and we sould account te comressblty o te lqud tsel. ercent I ever or te gases, ts assumton cannot be made gases are comressble and te ressure tey reduce te volume. Ten we assume te relatve velocty n drecton tat to smly t. Because, we can suermose to searate equatons, but, eort wll ncreasee n ts stuatons, n tose cases. Fnally, we assume two assumtons; te rgd surace wc n most o te case s true even or rubber materal stretcng does not aen n tat muc. So, tat s a reasonably good assumton. And nally, we say tat only te nclned surace squeee to smly te stuaton. But, weter te soluton we can get or drecton and drecton. And you can suer mose. So, tat s not a very bad assumton, t can be made and we can get good results. (Reer Slde Tme: 6:50) ressure Dstrbuton Dmenson n -dr s muc larger tan dmenson n -dr 6η U t, (, Z ) ( ) ( Dmenson n -dr s muc larger tan dmenson n -dr ) Now, tere s some nterest to tnk over. Wat ts equaton s really martng, wat ts equaton really conveyng to us to understand tat, wat we wll do wll non dmensonale

8 ts equaton. Wat s te reason or te non dmensonalng? On ts slde s gong to resent now, let us take two assumtons n one case, drecton dmensons are muc larger comared to te drecton, wle n second case drecton term or dmenson n drecton s muc smaller tan drecton, tey are two etreme cases. In one case we are assumng te drecton, dmenson s muc larger comared to dmenson n drecton. In second gure, we are assumng drecton dmenson s muc smaller comared to dmenson n drecton. So, we can wrte te dmenson n drecton s muc larger tan dmenson n drecton and te dstrbuton comes sometng lke ts. Now, wat we are ganng rom ts. You are sayng tat ressure mamum, ressure remans same. As value s more by wll be lesser, wle n ts case s a low remans same by wll be low ger tan by let I asked to neglect one o ts term I can I say tat value o s muc larger. So, by wll be smaller n tat case ts term can be neglected. Wle n oter case, wle we ave sort lengt n drecton and by wll be tere. So, n tat cases second term can be neglected. We can assume tat based on dmensons, we can get good results. To understand ts better let us less non-dmensonale. Wat s te reason or non dmensonalaton say wat s te mamum value o? Mamum dmenson n drecton assume tat s a catal. Tat means, ts wll vary rom 0 to mamum or bar value wll be varyng rom 0 to. Second term same tng mamum value o or you say dmenson n drecton, mamum value s same, ts s a varable, t wll vary rom 0 to catal n tat stuaton bar wll be 0. So, wat s te advantage o non dmensonalaton, n resect to al eresson, we are tryng to use te rst term, were te wll bar wll vary rom 0 to. Smlarly, bar wll vary rom 0 to one. So, we can do order analyss wat wll be multlcaton terms comes tat wll decde, wc term s gong to be domnatng. To get te meanngul results, we wll also non dmensonale lm tckness say lm tckness s a dvde. We say non dmenson o lm tckness o bar s lm tckness at any locaton dvde by clearance between two suraces s a mean clearance. I mamum and mnmum value are gven, we can average out we can get te results to some meanngul. Now we go aead wt, ts knd o non dmensonalaton, wle n ts case

9 lke a nstead o, wrte I wrte bar nto catal. Smlarly, nstead o, I wrte bar nto c. Smlarly, nstead o we can wrte bar nto and substtute over ere. So, ts s te same tng nstead o, we are wrtng bar nto catal nstead o wrtng. We are wrtng bar nto c, ere nstead o, you are wrtng bar nto catal tat s or te rst term. And wen we rearrange, wat we get term bar bar bar over ere and multlcaton actor comes as a c q by catal square. (Reer Slde Tme: :0) Ts s wat I was talkng about ts multlcaton actor s gong to gve us order analyss. I we do smlar knd or te drecton, wat we get, c by catal square. So, n ts case we are gettng smlar knd o term only, te non dmensonal orm ecet te n s stll n dmensonal orm, multlcaton actor comes as te c cube by catal square. Wle n second term s te c cube c cube s common n bot te term wle nstead o catal square n second term s comng catal square. omng to ts sde, wc we ave normaled as well as we are gettng multlcaton actor as a catal c dvde by catal. Now ere 6η s term and tere s nterest to non dmensonalse all tese terms. We already non dmensonaled, non dmensonaled, non dmensonaled ts, but, ressure s stll n a dmensonal orm. So, tere s a need to non dmensonale ts ressure. So, one way to non dmensonale, t s takng U as a common o ts s a 6η nto U, wle ere t wll be wtout U and ere t wll be dvdng c dvde by U, tat s comng over ere t U Z 6 η t U Z U U 6 6 η η tu t t Z U,, 6 η t 00 0, Z t , Z

10 c dvde by U. Tat means, U as been taken common rom ts bracket, te wole 6η nto U as been sted to te let and sde. Tat s rgt ere te c remans same catal square and 6η U s added ts s ressure unt. Ts s m m q. Ts s mm square o n te meter cube meter square. So, meter remans over ere, wle velocty s eressed n a meter er second. So, meter meter wll be cancelled and eta s generally ressure unt nto second. Second s over ere and dvde by second, tat wll be cancel out. So, overall ts eresson wll be er one dvde by ascal, one dvde by ressure unt. And wc means we multly ts ull c nto dvde by 6η U catal square, t wll turn out to be non dmensonal ressure term. So, tat s ow rearranged. In suc a manner, ts equaton can be rewrtten n ts orm say bar s a non dmensonal term, bar s non dmensonal term, bar s non dmensonal term, bar s non dmensonal term. Now, n ts stuaton watever te dmenson we coose, equaton wll reman same, soluton rocedure wll reman same, only te magntude wll cange. Tat wll brng a lot o stablty n numercal analyss, wc we can eress later. However, wat s te better tng comng over ere s n ts case, we are non dmensonalng wt a catal square term. And ere we ave a catal square term n tat to brng tat catal square term; we ave dvded as well as multled. So, ts s gong to make a non dmensonal ressure term, wle ts term remans, ere s a catal square, dvde by catal square. Ts s ressure unt non dmensonal ressure unt, were wll vary rom 0 to ere ressure term agan, were wll vary rom 0 to. And mamum value o te ressure wll reman same, n drecton and drecton te same ressure mamum value. So, tey can be treated as same unt may be say overall ressure term, comes rom as a one unt, one same ts wll come also unt one. But, ts s square term by s square s gong to make derence. Agan say ts s gong to gve me te quattatve analyss, weter second term s gong to domnate or rst term s gong to domnate. I one term s domnatng muc larger etent comared to oter, I can neglect one term. Tat wll smly soluton to greater etent ever n ts term, we not non dmensonale te tme. Wc was mortant also because to get all non dmensonal term s also mortant to nondmensonale tme. Tat s wy we dd U nto t dvde by c at te meter er second nto

11 second dvde by meter tat s turnng out to be n non dmensonal, tat s non dmensonal tme term. So, we ave to non dmensonale ts comlete equaton. As I mentoned ts equaton s gong to gve us order analyss a quanttatve results also. So, we are gong to see tat now assumton s a, dmenson n drecton s 0 tmes comared to dmenson n drecton or we say by s 0. I substtute over ere, wat we are gettng ere s 00, ts s a unt, ts second term s only unt, t as been multled wt te 00. Tat means, ts term s very rc comared to rst term or even you neglect ts term I am gong to get an accuracy roblem u to ercent. I am gong to get 99 ercent accurate result tat s advsable. We need to see, ow muc tme really requre to solve ts equaton and I am gettng even neglectng ts term. I am gettng 99 ercent ostve results and tere s no ont or tere s no meanng to account ts relaton, unless I am requrng accuracy to muc larger etent. Wle te second and ts source terms are remanng te same, tere s a non dmensonal terms. Smlarly, I assume te reverse u ts, we n ts case, we say tat ts term wll be neglgble. Tat s very clear ater non-dmensonale beore a non dmensonalaton. It was not very clear ater non dmensonalaton, s very clear tat ts tme s neglgble catal by catal s 0 or dmenson n drecton s 0 tmes comared to dmensons n drecton, reverse s also ossble. You say tat s te muc smaller comared to. In tat stuaton we substtute over ere I am gettng 0.. Tat means, wen ts dmenson s muc smaller, dmenson s muc smaller, we are gettng low value o. Tat was eressed n revous slde wt a guratve orm. Wen dmenson s muc larger, ten only ts term s meanngul, rst term s not carryng muc wegt. Here te 00t wegt comared to one wegt or we are gong to get 99 ercent accurate result, even we neglect te rst term. Wle n second term, ts s neglgble, ts s gettng one way wle ts second term s gettng only ercent o tat. Even I tnk n ts case I neglect ts term, I am gong to get accurate results uto 99 ercent. So, we can say ts term s neglgble. Ts knd o analyss s gong to el us, to solve te, ts equaton. Solvng artal derental equaton requres some eort, requre some numercal metod to be mled. And numercal metod wenever we mly, we need to tnk ow to drect tat soluton.

12 Numercal solutons are not easy to solve wtout understandng t. I we kee gvng nstructon to te comuter to solve te equaton, wtout understandng tat, we wll not be able to get more meanngul results. Tat s wy I am emasng to understand system rst, understand te dmenson rst and ten you tnk wc numercal analyss sould be aled. I t does not matter muc your tme, does not matter muc you can go aead. But, tme matters to us ten we sould te tnk soluton based on dmensons. (Reer Slde Tme: 0:9) Now, tere s one moded verson also you say tat we are assumng tat, tere s a steady state Reynolds equaton. lates are not gong to cange te oston wt resect to tme. Tat remans nclnaton, remans same lm tckness, and remans same wt resect to tme. So, n tat case ts second term say you can source term wll be neglgble and tat wll be called a steady state Reynolds equaton. So, ts term s neglgble. Now, ts s wat we are talkng, load s aled, velocty o te yellow late along te drecton tat s u velocty. And ts late s a statonary and relate to lm tckness between ts two lates at any oston s remanng ed, s not gong to be canged wt tme. Anyway ts s a roved assumton we know very well because o te condton some t Z ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ 0.5, 0.5,,,,,, ( ),,,, Δ

13 cange wll occur, but tat may not be muc roblem. We can do some sort o a statstcal analyss. We can assume te ressure wll vary. So, or startng urose s always advsable to start wt te steady state analyss, unless turn te ressure generaton. And see ow te results are, we are able to get wat knd o modcaton n desgn sould mlement, so tat we can get muc more desrable results. Now, ts s sown n te two lates and we want to tnk about solvng ts. We can use some numercal metod, one o te numercal metod s te nal derence metod. Wat we do n a nal derence? We dvde ull surace n number o nodes. You say bg one cannot be taken easly, we dvde te one n number o sub ones, ten tacklng eac sub one wll be easer comared to tacklng one bg one. Tat s wy we dvde te surace n sub number o nodes tat s wy n drecton and drecton. Tese lnes are arallel lnes and erendcular lnes, tey are gong to gve us some nodes. And wat we are sayng tat we are nterested to nd te ressure at some nodes, assume ts s, uncton and we are takng te el o negbourng nodes to nd out te solutons. And t s teratve rocedure. I am gong to elan ow t s teratve rocedure and ow ts tese negbourng nodes are gong to el us? Wen we talk to dvde ts wole surace n a sub number o ones, we say tat tese are te nodes ntersecton onts o te nodes. Were te ressure wll be calculated and at any, wen I cannot ts equaton I need to gve some nodes. I say I can vary rom, number o nodes total number o nodes n drecton and can vary rom one number o node to number o node n drecton. Tat s bar, and tere s a ressure term, ressure gradent term tat can be reresent n nal derence orm sometng lke ts. I we are sayng tat, we are averagng term. We are makng te average o ts to nd out te, gradent at te,, we are takng te el o advance node n drecton and node wc s lesser tan by unt comared to,. Tat s wy te, and -,. I I know never nodes value, I can get te results and we go or te second gradent, n tat case we requred to calculate at te md node, ten tere s somewere ere n between some ressure gradent at te md node. Because, we know ts can be ater a mdterm wll not be tere, md node wll not be tere, wen we mlement ts ormula over ere wll be te only te nodes.

14 So, tat s te some sort o accuracy requrement, we arrange t. We can elan ts wt a Taylor s seres. We know, te Taylors seres s dened and tere s some delta ncrement, tat Taylors seres can be eressed n terms o lus delta. And rst gradent o, rst gradent o wt lus to second term. Smlarly, wen we take slgt derence n negatve sgn, say mnus delta delta tat also can be reresented as mnus delta. And ts gradent (derentaton o wt resect to ) can be gven and ts wll be te second ordered term. As our assumton ts delta s very small ten ger terms can be neglected. Now, I subtract second equaton rom te rst equaton. Wat I get o lus delta mnus o mnus delta s equal to nto del and ressure gradent. And rearrangng ts, we can nd te gradent o n drecton as lus o lus delta mnus o mnus delta dvde by delta. And tere s te same tng wc we ave eressed over ere lus ressure. And watever I am assumng te at any ont s te, watever te net delta come, watever te net derence comes tat s te lus. Tat means, te dstance between, tese onts s equal to delta so, lus and mnus. Tat means, I am calculatng te ere, I am takng ts term mnus ts term dvde by dstance between tese terms as a nal derence term. We are calculatng an accuracy s reasonable, we assume te delta s a smaller, t s not gong to aect ts soluton accuracy. So, based on tat, we can really dvde ts equaton n number o algebrac equatons. And we know very well soluton o te algebrac equaton s muc muc aster comared to artal derental equatons. Now we can go sequentally, we can go smultaneously. Tere are number o metods avalable or tme beng I am ust tnkng about te sequentally to elan n ts analogy o neglectng term comared to oter term usng sequental stes. And ts equaton ater substtutng value, wat we ave aceved rom ts over ere we can rearrange and we can get ts eressons tat s t lus mnus lus mnus dvde by del square. So, we know ow to rewrte ts artal derental equaton n terms o nal derence, n terms o te nodal ressures. As nodal ressure can be eressed n nodal ressure and we nally, want te nodal ressure to be evaluated. So, we requre some sort o metod wc we are gong to elan n te net slde, ow te equatons gong to el us?

15 (Reer Slde Tme: 8:7) Now, ts say tat we ave tree terms, n Reynolds equaton tat s steady state. Reynolds equaton wt a number o assumton, te rst term can be te resent ere s te only gradent. We wll take te al node calculatng te and te al node s not a roblem s a geometry and wc s well dened to us. Wle, ressure s te second gradent tat s wy t s comng lus lus agan lm tckness at te al node, mnus al node and ressure wll be mnus. Well trd term s te submsson o ts n revous slde, we wrote t wle n ts case tere s a submsson. So, tey ave bot as same, t wll turn out to be and dvded by s square. Because ere ts s a velos s turnng term s comng, tat s wy we ave wrtten wole term lke ts. Same tng or te s comng we can wole term sometng lke ts ere nstead o lus 0.5 we are wrtng c, but, ere we are wrtng lus ont ve wle ere t was only. Tat means, varaton n drecton and drecton was consdered. Wle n ts case varaton n te I drecton s not consdered varaton only n drecton s beng consdered, ts s wat we eress usng te nal derence metod. Smlarly, or trd we can smly wrte, nal derence lus mnus te ressure gradent sorry lm tckness gradent s only n drecton. So, we can wrte te comlete te Reynolds equaton n terms o nodal ressures and geometrc terms. We can rearrange n muc more meanngul reasonable way tat s lke ts. So, we want nodal ressure at any node tat wll be a uncton o nodal ressure at lus. Smlarly, t wll be a uncton o mnus uncton o lus and uncton o te mnus.

16 You can see ts s te centre ont, we ave negatve drecton, ostve drecton, negatve drecton and ostve drecton. Tat means, we are calculatng te low ressure around te nodal ressure, were we want calculaton and based on tat we are evaluatng te results. It s more lke an averagng a sceme, I know ressure term nearby and try to gure out wat wll be ressure at te nodal ont o tat. In addton, tere wll be a constant term all tese ressures are 0. Intally ts term s gong to decde wat wll be. Ts s a wat we ave eressed over ere, I, need a s somewere ere takng result o mnus. I am takng results o lus, I am takng te result o mnus and I am takng te result o lus to evaluate wat wll be and tese are te al node onts. Weter we need to nd out wat wll be te lm tckness? As I say te lm tckness calculaton s not a roblem because, t s a geometry deendent and n a geometry s gven a contnuous uncton. We ust need to substtute te value o to nd out wat wll be te at tat locaton? See t s not dcult, t can be gure out and once we nd, we do not ave to do any teratons to vary tat results. But, ressure we need to nd out tat s requrng t to te rocedure. Just to gve eamle, we can take a. We wll say assume te node ont s somewere ere s gong to be a uncton o uncton o 0 uncton o and 0 ts s a generally boundary values. n te ntal ont and te n te ntal ont were te 0 0 term s eressed generally, tese are te boundary values. I and 99.9 ercent we know te results o te boundares generally at te edges or te startng onts, we wll be knowng wat wll be te ressure value. So, we can say n ts stuaton, we wll be knowng wat wll be te 0 and wat wll be te 0 n most o te cases and wat we need to nd out te net one. As ts term s te and we are tryng to nd out and we are gettng te value rom and wc are not known to us. We are eectng results rom uneected tngs. So, tat s wy we requre teratons, we need sequence o teratons. So, tat we can nally, come to converson ont ntally, we can assume ressures are 0, and only ts term s gong to gve me te results. An nterestng ont s tat, I ave wrtten b c d e. Ts s say tat deend on te geometry, deend on te sacng one, ow muc sacng we requre between two node based

17 on tat. Once t as been decded, ts value remans ed t wll tese values are not gong to cange wt te teratons, wt teraton only te ressure terms are gong to cange. Now, ts s wat I elaned and I mentoned about te teraton. It s sceme also we say tat we want ressure at node or may be n ts case we took eamle o. So, we need at a any teraton k lus assumng te k teraton s already over. We evaluate ressure at te k lus teraton t wll be uncton o, ere te teraton s k. Tat means, revous teraton weter te ressure known to us tat s gong to el us to nd out ressure at a node. And ere agan 0 comes wc we already ave overcome tat s a revous node comared to ts node, wc wll be known to us. Tat s wy ere we are wrtng k lus teraton. So, watever ressure s already known, we can drectly utle. Wle comng to ts term ere agan wll not be known beore and and or k teraton. So, we take a eamle, we take a value wc was avalable rom te revous teraton and comng to net term d and 0 ere agan tat 0 s already over we already assed ts to net node we are advancng. So, ts value o ts ressure wll be known to us. So, t wll be k lus tat s recent teraton. Tat means, terms are comng rom revous teratons and terms are comng at te resent teraton. Tat s gong to decde us, decde our ressure term wle ts term wll reman constant or we say all te a b c d e wll reman constant, cannot vary wt teraton. So, teratons are only requred or ts urose. And or smlcty, wen we start soluton, we assume ressure at all te nodes s equal to 0 and we can assume tat as a k equal to 0 te 0 teraton. We are terated anytng ust begnnng. So, te 0 teraton ressures are known tat s equal to value equal to 0 tat s known to us. Now, we can start uttng all te at ts e wll gve us, wat wll be te value o ressure at node or te rst teraton? So, tat wll gve a sequence tat wll gve good result to us.

18 (Reer Slde Tme: 46:8) n m, teraton k n m, n m teraton k, teraton k- ε Now, te queston comes ow many calculatons are really requred, ow many teratons are really requred to nd out; wat wll be te ressure? Say ts s a reettve rocess, we need to reeat tll we get desrable accuracy. It may be ercent o te ressure, 0. ercent o ressure, 0.0 ercent o ressure, deends on. Wat s te order o analyss are we dong nal desgn or we are dong te ntal desgn, by ntal desgn you may not requre. So, many teratons we requre some ntal geometry. So, tat we can come u wt te some results and ten go wt te comatblty wt te oter comonents. So, wat we say ressure conversons term s based on te ressure, wat s te meanng o ts? We say tat all te nodal ressures we are summng u. Wat s te value o all te nodal ressures? For cavtaton key, wc s te resent teraton suose you are and ts s te revous teraton and dvde wt te submsson on absolute value. We are talkng only te absolute value, may be sometme negatve value also comes, ave to be go aead wt sort o cavtaton wt negatve ressure s generated or you say ressure glow te atmoserc ressure s generated. So, overall ts ercentage sould be lesser tan some eslon value, t can be 0.0 t can be 0.00, deends on te requrement. Now I am comng to te ont, wc we started, ow non dmensonaton s elng us to nd wc term s gong to domnate?

19 And I know wc term s gong to domnate, ow we are gong to save te calculaton, ow we are able to save te eorts, to llustrate tat? We are gvng ts say, assumng I dvde t lates n drecton n 5 nodes. Smlarly, n drecton we dvde t n 5 nodes and we requre 0 teratons to come u wt te rgt soluton. It s ust an assumton, ust to eamle to convnce wat s te meanng o neglectng one term and ow muc benet we are gong to get. So, ow many stes are requred 5 nto 5 nto 0, te assumtons are tese all te stes are requred to nd out te ressure varaton n trbal surace. Tat s gong to ts gure turn out to be 8,750. Now, I take ts assumton, say dmenson n drecton s 0 tmes dmenson n drecton. Wat we are gong to gan sometng lke ts, 00versus, ts term s neglected. So, I ad to consder only n drecton, wat s te meanng o tat I do not requre any node n drecton. Tey sould term n drectons are neglgble. Wy do we requre any node I do not ave to dvde tat late? Assume te ressure wll be known to me. And ressure n twenty 5 same number o nodes we are keeng n drecton as we are not countng te ressure term n n drecton. We can nd ressure n, sorry, terms wll or teratons wll be almost al o te term teratons, wc we requre or te bot te drectons. We say wen te ressure term s unknown n drecton as well as drecton, we are requrng 0 teratons. Wen ressure s not calculated n drecton at all s calculated only n drecton and keeng te same number o nodes. Tere s a ossblty we requre only 50 ercent o teratons comared to ts and nstead o 0 we requred only 5 teratons. So, I multly 5 nto 5 tat s gong to gve me 75. You can come ere 8750 was a 75 calculatons. And we know te stes and tese are te stes and eac stes tere wll be number o calculatons, may be more tan 000 calculaton o eac ste. So, ow many stes we are gong to save rom ts? Is almost a 50 tmes or you say dvde by 75 s gong to gve me 50 gure. I mss better understandng ts system, we are able to save 50 tmes eorts or we assume tat eac eort s requrng 000 calculaton. Tat means, we are gong to save calculatons. Wc may be a reasonable gure artcularly, wen we are startng te soluton sceme or we learnng te trbology subect. In tose stuatons s gong to really el us to a greater etent tat s useul tng or us.

20 So, ts s wat we eress, ow we get te benet o te non dmensonalaton and wen we non-dmensonale ts result, ow we are gong to get overall calculaton benets? (Reer Slde Tme: 5:9) 0., Z 0.0 I wll cover n net lecture, but, or te tme beng I can sow wat we eressed n revous slde eamle related to tat. I am assumng tere s a rst load or load erendcular to te as s comng over t. Tere s late wt statonery late, sldng late wc as a velocty u and ts velocty n te resent eamle s gven 0 meters er second, wc s very, very g velocty. And we are sayng te lm tckness o te two unctons are gven. At te entrance, te lm tckness s 40 mcron and at te et lm tckness s 0 mcron. Ts s te reasonable gure assumng tat wen te dmensons are 0mm and 00mm, ts lm tckness s reasonable 40 to 0. And wat s vscosty o te ol wc we use? We are usng sometng lke a 0 mllascal. Second, now nterestng ont s tat we are dened te dmenson say te yellow late dmenson n drecton s 0mm, wle n drecton s a 00 mm. Tat means, by s gong to turn out to be onts because, s only 0 unt and s 00 unts. So, 0 by 00 wll be turn out to be ont one tat means, we multly, wt a square by square, net s te second term wll be neglected. Second term s neglgble, artcularly n ts stuaton. So, tat wll smly our soluton say by s ont one. So, ts term wll be neglected and ts term s gong to be only useul or

21 us. Based on tat we can solve ts, we can use a nal derence to solve t. I wll cover ts n my net lecture. Tanks or your attenton.