.5 Final Exam 005 SOLUTION Question U A Coss Section Photo emoved fo copyight easons. Souce: Figue 7l in Clanet, C. "Dynamics and stability of wate bells." J. Fluid Mech 40 (00): -47. R d Tooidal im U Two identical jets of wate impact on one anothe foming a sheet of liquid as shown in the figue (and as shown in the suface tension movie). The sheet gows to a adius R ending in a tooidal im that ejects doplets of fluid. Conside incoming jet velocities, U = m/s, and coss-sectional jet aeas, A = π/4 cm.. Calculate the Reynolds numbe, Capillay numbe, and Foude numbe fo this flow. Use these numbes to ague which effects (e.g. inetia, viscosity, gavity, etc.) ae impotant in this system. Solution: U = 00 cm/s, A = π/4 cm D = cm, µ = 0 dyne-s/cm, ρ = gm/cm, σ = 70 dynes/cm. Hence: Re = Ca = F = ρud 0 4 inetia viscosity µ () µu 0 suface tension viscosity σ () U 0 inetia gavity gd () If the sheet is flat, we can neglect gavity (if the sheet is cuved, as in pat 4, we must include gavity).. Deive an expession fo the maximum adius R of the liquid sheet and calculate R using the values given above. Solution: The maximum adius occus when the sheet is flat so we neglect gavity. Apply consevation of momentum on the following contol volume:
σ σ By Benoulli, u = u = U. Consevation of momentum yields (note: p = p a ): d ρv dv + ρv(v v c ) n da = pn da σdl dt CV CS CS CL (4) ρu πd = σπ (5) whee d = d() is the thickness of the sheet. Thus inetia balances suface tension at steady state when σ d =. (6) ρu By consevation of mass ( on the LHS fom two incoming jets): Combining this with (6) we find A UA =πdu =. (7) πd ρu A = R im = 8 cm (fo wate) (8) πσ. Using enegy aguments, show that the tooidal im is unstable and will beak into doplets. Deive a lowe bound fo the size of the ejected doplets. Solution: E suf = σa suf (9) Thus, if the tous has a lage suface aea that the doplets (whee the volume of fluid is identical in both configuations), suface tension will beak the tous into dops. The tous is unstable if 4 V 0(πR im ) = V dop = N π tous = πr dop (0) N = R 0 (πr im) () 4 dop S tous πr 0 (πr im ) > S dop = N4πdop () dop > R 0 () Thus the tous will beak into doplets with adii > R 0 to lowe the suface enegy. (Note, tiny doplets ae no good since they ceate exta suface aea and hence cost moe enegy.) To calculate the actual dop adius athe than a lowe bound, we need to conside how to move fluid fom the tous into the doplets. This calculation is a bit moe involved and esults in a citical wavelength of 9.0R 0 (see Rayleigh-Plateau instability).
4. If gavity becomes impotant in the poblem, the sheet will sag and become a wate bell (instead of a flat sheet) as sketched below. Show that the shape of the bell is given by the solution to the following equation whee s is the coodinate along the sheet and othe vaiables ae defined in the sketch: ( ) πσ U +gz dθ sin θ dθ ( ) ρg cos θ + + = ρ U +gz UA ds ds whee s, z and θ ae elated by dz/ds = sin θ. You do NOT need to solve this equation! Solution: y θ(s) x s g θ(s+ s) s+ s Consevation of momentum: d ρv dv + ρv(v v c ) n da = pn da + ρg dv dt CV CS CV + σt y sin θ σt y dl dl + σ da (4) CL s CL s+ s CS } {{} out-of-plane cuvatue In the y diection at steady state: sin θ ρu πd sin( θ) = ρgπd s cos θ +σπ sin( θ)+σ π s (5) In the limit s 0, sin( θ) θ and θ s s. Hence ( ) σ θ sin θ ρu θ = ρg cos θ + + s d s Fom Benoulli: Fom consevation of mass: Combining (6), (7) and (8) we find ( ) πσ U +gz dθ sin θ dθ ( ) ρg cos θ + + = ρ U +gz (9) UA ds ds θ (6) u = U +gz (7) UA =πdu d = A π (8)
.5 Fall 005 Final Exam Solutions to Question : Flow Focusing in Micofluidics A) Dimensional Analysis [an easy 0 points!] i) We seek a solution which descibes R j = f( H,,µ o,,q o,,, ) Se we have n = 9, = (MLT) and hence n = 6. Most of these can be found by inspection. Picking and H to chaacteize fluid, flow and geomety espectively we find: R j Q 0 =, µ 0, H,, H H whee Ca i µ( H ) is a capillay numbe and Re i H is the Reynolds numbe fo the inne fluid. Note that (angle in adians) is aleady dimensionless and can be vaied independently of H, so it is also a dimensionless goup!! ii) We need to find a poduct goup involving suface tension, viscosity and density that has units of length: a c a i.e. l c µ b o equivalently a dimensionless goup: = H b c Since we have unknowns (a, b, c) and equations thee is one unique solution: l c µ This is often efeed to now as the Egges length (J. Egges, Phys. Rev. Lett.,7 (99). The coesponding dimensionless goup is the Ohnesoge numbe l = c Oh = ( ) H i and it measues elative impotance of viscous effects to inetial and capillay effects in jet beakup ( 0 ) Fo wate: l = c =.9 nm; Fo glyceol: l = ( 0 )( 7 0 c ) ( 0 )( 6 0 ) = 6mm Note that even in a micofluidic geomety (with say H = 0 m) the beakup of a wate jet is not affected by viscosity until scales of 0(0-0 nm)!! In contast fo glyceol we should expect viscosity to dominate all aspects of the pocess. B. Conveging Plana Channel i) the continuity equation in cylindical coodinates is
.5 Fall 005 Final Exam Substituting fo v =! f (" ) we find!( v ) +!v "!!" +!v z!z = 0! %! " # f ($) ( & ' ) * + 0 + 0 = 0 The volumetic flow ate (pe unit depth) though the device at any distance is: Note that the function f! of the flow ate into the thoat). ii) +$! = % v " ( e ) & e d" = % f " d" = constant! (B) #$ ( ) has units of m / s If we substitute v =! f ("),0,o!p! = µ )! " +! # $ * + +$ #$ ( )!" # $ and! is positive hee (i.e. it is the magnitude [ ] T into the Navie-Stokes equations we obtain ( )!! v % & ' +! v,.!( -. / 0v!v! and!p!" = µ Substituting fo v = f (!), we see that the fist tem on the ight hand side is zeo and hence we obtain:!p! = " µ f ## + $ f Similaly fo the! component we obtain:!p!" = # µ f $ whee a pime indicates a deivative with espect to theta.!v!" (B) (B) iii) If we take coss-deivatives of B and B we obtain:! #!p &!" $ %! ' ( = ) f *** + + f f * and! #!p &! $ %!" ' ( = + 4 f ) Equating and then eaanging by multiplying by (! µ ) gives: f!!! " # µ f f! + 4 f! = 0 o equivalently if we define a dimensionless function f * o F == f! then we obtain: d F d! " % # Q$ ( & ' ) * F df d! + 4 F $ = 0 (B4) The dimensionless goup! = " # is the Reynolds numbe fo this poblem (as we found in pat A). The appopiate bounday conditions fo this poblem ae a) no slip at the walls: at! = ±" ; v = 0 so hence f (±!) = 0
.5 Fall 005 Final Exam iv) b) this gives two bounday conditions but (B4) is a thid ode ODE so we need anothe bounday condition. A statement of symmety at the centeline is NOT sufficient as it is tantamount to the same thing as (a). Howeve we can also note + that the volumetic flow ate though the device is constant o: F() d =. In the limit 0, the nonlinea d ode ODE (B4) becomes a linea ODE. You should ecognize that this is easy to solve in tems of tig. Functions (hint; you can nd diectly integate it one to find a ode ODE and find the complementay function and paticula integal). Regadless, we ae given F() = A + B sin + C cos and hence we immediately find: F = B cos C sin F = 4 B sin 4C cos F + 4F = 0 F = 8B cos + 8C sin Applying the bounday condition (a) we obtain: B = 0, A = C cos. To detemine A and C we now use the final bounday condition: + F()d = C cos C cos d = C = + and the final velocity field is: { } sin cos [ ] f () Q cos = i cos cos v = = sin cos [tan ] cos A quick sign check shows that this is inwad eveywhee, has geatest magnitude at the centeline ( = 0 ) and goes smoothly to zeo at the wall =±. p f f v) The adial pessue gadient is (fom B): = µ +. The second tem is obviously positive. Also fom the fom of the velocity field we can see that the second deivative of the function f ()~ cos is f ~ 4 cos and thus the fist tem also inceases with. So the conclusion is the pessue always inceases away fom the apex ( = 0) both viscous effects and inetial effects (Benoulli) lead to a pessue dop as the fluid acceleates into the contaction thoat. C. Axisymmetic Potential Flow Into A Capillay Tube (i) m We ae given the velocity potential = cos and thus the velocity field is given by m cos m sin v = = and v = =.
.5 Fall 005 Final Exam Note that thee ae TWO components so the velocity vecto is not puely adial in this case. Also note that it decays away as (unlike the viscous flow solution above). Finally note that m has units [m 4 /s]. The steamtubes can be found fom eithe expession fo v o v to be: m cos m sin = (C) sin = v = Note that the steamfunction is only defined to within a constant also note it has units of [m /s]. To daw this in the {R,z} plane we can substitute fo z = cos, R = sin and by Pythagoas = R + z, sin = R (R + z ) to find mr o equivalently ( m) = (z + R ) = R 4 (R + z ) / Given a doublet of stength m this is an implicit equation fo a steamline R i (z) fo any value of and could be plotted in Matlab. Altenately, we can see that the flow is focused down and gets pogessively thinne so we can eaange to give: o simplifying it fo small values of (ii) ( m) 6 z ( + (R i / z) 4 ) = R i Rz << : / = Kz / R i ±( m)z The pessue field is given by the Benoulli equation. Fa away fom the tube the velocity decays to zeo as and the pessue is p. Hence we find p v v = p(,) + + 0 = p(,) + {v + v } (C) (iii) o m p(,) = p m 6 {sin + 4cos } = p { + cos } (C) 6 At the inteface, the shea stesses balance and the nomal stesses only diffe by the additional pessue associated with cuvatue. Fomally (if n is a nomal vecto and t is a tangent vecto): (i) (o) (i) tn = tn = p (o) (o) and p (i) + nn + nn + H whee the mean cuvatue fo this axisymmetic jet is H R i R i. Also note that we equie the velocity vectos to match; v i = v o. It thus looks like we have FOUR nd bounday conditions at the inteface (which we do we have to solve a ode ODE fo each fluid!) (iv) As we have shown in C, the dividing steamline is a cuve of the fom R i = Kz / (in cylindical polas). If we wite down the lubication equations, we obtain 4
.5 Fall 005 Final Exam v z p 4m 0 = p + with solution v = R z = p R + c R + c whee z µ z z 7 Howeve when we ty and match the velocity at the inteface thee is no slope since Rz (i) = 0 = µ v z. Similaly thee is no slope at the centeline R = 0 and the solution is thus a i simple plug flow with m cos m m sin v = = and v = = 0. z Note that the angle tan = R z = Kz / 0, so the flow is inceasingly close to a lubication flow. Although not asked to check this, you can see that this solution also satisfies consevation of mass because: m z = R j v = (Kz / ) ( ) = const. (v) Conside a coodinate system (x,y) that is locally nomal to the inteface and in the flow diection R = R i (z) ; i.e. x z and y = R R i (z). The bounday laye equations ae then (neglecting g): v x v x p (o) v x p (o) v x + v y = + and 0 = x y x y y whee p (o) is the pessue field fo inviscid flow which is imposed on the bounday laye and which we found in pat (ii) see eq. C. The solution to this equation would be diffeent than the bounday laye solution fo two easons:. Thee is a favoable pessue gadient; the pessue deceases as we focus the flow towads the inlet; with the gadient given by: p (o) dp (o) 4m = = x dz z 7. The appopiate bounday condition to apply at the bounday is NOT v x (y = 0) = 0 but athe the no-tangential stess condition: w = µ y x Both of these effects ae likely to make the bounday laye gow much moe slowly than the Blausius solution. v y=0 = 0. 5