2.25 Advanced Fluid Mechanics

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1 MIT Depatment of Mechanical Engineeing.5 Advanced Fluid Mechanics Poblem 6.1 This poblem is fom Advanced Fluid Mechanics Poblems by A.H. Shapio and A.A. Sonin The sketch shows a cicula beaing pad which ests on a flat base though the intemediay of a film of viscous liquid of instantaneous thickness h(t). The load W causes the pad to sink slowly at the speed S, and this squeezes the liquid out fom unde the pad. Assume that h «D, that the viscosity is vey high, and that the speed S is vey small. (a) Making appoximations (state them pecisely) consistent with theses assumptions, show that the settling speed is 3 W h 3 S =. (6.1a) 3π µd 4 (b) An appaatus with two vey flat plates of.3 m diamete caies a load of 1 kg on a film.3 cm gm cm thick. If the liquid is a heavy oil with a kinematic viscosity of 1 s and a density of.93 cm3, estimate the speed S. (c) If the load W is constant, and the gap wih is h at time zeo, show that the wih h vaies with time accodingly to 1 h 64 W h = 1 + t. (6.1b) h 3π µd 4 (d) Calculate, fo the initial conditions of pat (b), the time (in hous) equied fo the gap wih to be deceased to half its initial value. (e) Suppose now that the initial thickness is h, and that a constant upwad foce F pulls the disk away fom the base. Show that the disk will be pulled away in a time 3π µd 4 t =. (6.1c) 64 h F NB When h is vey small, the time t is vey lage. This is the basis fo the phenomenon of viscous adhesion, e.g., adhesives such as Scotch tape, o the appaent adhesion of accuately-gound metal sufaces..5 Advanced Fluid Mechanics 1 c 1, MIT

2 Viscous Flows A.H. Shapio and A.A. Sonin 6.1 Solution: (a) In tems of the flow geomety, the poblem is simila to 6.3; the diffeence hee being that the flow is unsteady. Fist, let s use the continuity equation to get some infomation about the ode of magnitude of the velocities, 1 1 u z u u z (u ) + =,. (6.1d) z z Now, let s go to the NS equations. Witing the NS equations in cylindical gap between the beaing pad and gound, ( ) [ ( ) u u ] u θ u u uθ p 1 u 1 u u u u θ ρ + u + + u z = +µ + + aa t aa aa θ z aa θ z θ aa } aa aa aa aa I II = III IV = V V I = (6.1e) whee we elabeled the tems to futhe make an ode of magnitude estimation, also notice that due to symmety, the tems. Let s stat compaing the tems IV and V, θ IV h «1, (6.1f) V D so we can neglect the tem IV compaed with tem V. Also, notice that V I IV, then V I can be neglected when compaed to V. Now, let s compae II to V, then II ρv D h «1, (6.1g) V µ D then II can also be neglected when compaed to V. Now, when compaing III to V, we fist notice that II III since II u / u z /z 1, (6.1h) III u z /z u z /z whee we have used the infomation obtained fom mass consevation. Hence, III vanishes when compaed to V. Now, when compaing I to V, we have I ρh, (6.1i) V µτ whee τ is the time scale involved in the pocess. The souce of unsteadyness is the pad settling down, which endes u and othe flow vaiables time-dependent. Hence, h ρsh τ, (6.1j) S µ Since it is given that S is vey small and also h is small, we can safely assume that I can be neglected compaed to V. Hence we have the NS equation as 1 By now, you should be familia with this method of obtaining exta infomation that can be quite useful when compaing tems in the NS equations. We ll compae the tems with tem V because, since the gap is small, this is likely to be the lagest deivative..5 Advanced Fluid Mechanics c 1, MIT

3 Viscous Flows A.H. Shapio and A.A. Sonin 6.1 Integating we get u (z), p + µ u z =. (6.1k) u = 1 ( p ) (zh z ) µ (6.1l) h h 3 ( ) π p h 3 ( ) π dp Q() = π u (z)dz = = (6.1m) 6µ 6µ d Note that Q is a function of, since the settling down of the pad dives geate and geate flow ates as inceases! This can be veified by applying mass consevation in a cylindical CV, as shown. The height of this CV changes as h(t). ρdv + ρ(v V ) nda =, (6.1n) t CV CV cs (π h) + πhu =, π dh = πhu = Q(). t (6.1o) Since dh = S, we have Q() = π S. (6.1p) Now, we can find the value of the pessue gadient using the flow, then whee the BC used is p(d/) = (gauge pessue). dp 6µQ 6µπ S 6µS = = =, (6.1q) d h 3 π h 3 π h ( 6µS 3µS D ) dp = d, p() =, (6.1) h 3 h 3 4 Now, we can pefom a vetical foce balance on the pad, D 3µS ( D ) 3πµSD 4 W = πd =, (6.1s) h 3 4 3h 3.5 Advanced Fluid Mechanics 3 c 1, MIT

4 Viscous Flows A.H. Shapio and A.A. Sonin 6.1 then, we can finally get the velocity S = 3W h3 3πµD 4, (6.1t) (b) Plugging in the numbes, we obtain S = π vey small, which agees with ou assumptions. = [m/s] (6.1u) (c) Fom the velocity equation, we can integate to obtain the displacement S = dh = 3W h3 3πµD 4 h h dh ' h '3 = t 3W 3πµD 4 ', (6.1v) then, upon integation we have h t 1 1 3W t ' W t W h = = = 1 + t (6.1w) h ' 3πµD 4 h h 3πµD 4 h h 3πµD 4 h which gives [ 1 h 64W h ] = 1 + t. (6.1x) h 3πµD 4 (d) Using h = h, and plugging in the vaiables, we get the equied time as 1.4 hous. (e) Since S ' = dh, instead of dh, we have As the disk is pulled away, h then fo this limit, F t =. (6.1y) h h 3πµD 4 3πµD 4 t =. (6.1z) 64h F D Poblem Solution by MK/MC(Updated), Fall 8.5 Advanced Fluid Mechanics 4 c 1, MIT

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