Problem 3C - Stemjock
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 1 of 12Problem 3C.5Two-phase interfacial boundary conditions. In §2.1, boundary conditions for solving viscousflow problems were given. At that point no mention was made of the role of interfacial tension.At the interface between two immiscible fluids, I and II, the following boundary condition shouldbe used:7 11I IIIIIIII σ(3C.5-1)n (p p ) [n · (τ τ )] nR1 R2This is essentially a momentum balance written for an interfacial element dS with no matterpassing through it, and with no interfacial mass or viscosity. Here nI is the unit vector normal todS and pointing into phase I. The quantities R1 and R2 are the principal radii of curvature at dS,and each of these is positive if its center lies in phase I. The sum (1/R1 ) (1/R2 ) can also beexpressed as ( · nI ). The quantity σ is the interfacial tension, assumed constant.(a) Show that, for a spherical droplet of I at rest in a second medium II, Laplace’s equation 11IIIp p σ(3C.5-2)R1 R2relates the pressures inside and outside the droplet. Is the pressure in phase I greater thanthat in phase II, or the reverse? What is the relation between the pressures at a planarinterface?(b) Show that Eq. 3C.5-1 leads to the following dimensionless boundary condition. ρII ρIρI gl0 n (P P ) nhv02 II II I µρ11σµIIIIII[n · γ̇ ] [n · γ̇ ] I n (3C.5-3) l0 v0 ρIl0 v0 ρIIρl0 v02 ρIR1 R2IIIII in which h (h h0 )/l0 is the dimensionless elevation of dS, γ̇ I and γ̇ II are dimensionless rate-of-deformation tensors, and R1 R1 /l0 and R2 R2 /l0 are dimensionless radii ofcurvature. Furthermore PI P II pI p0 ρI g(h h0 )ρI v02pII p0 ρII g(h h0 )ρI v02(3C.5-4, 5)In the above, the zero-subscripted quantities are the scale factors, valid in both phases.Identify the dimensionless groups that appear in Eq. 3C.5-3.(c) Show how the result in (b) simplifies to Eq. 3.7-36 under the assumptions made in Example3.7-2.Solution7L. Landau and E. M. Lifshitz, Fluid Mechanics, Pergamon, Oxford, 2nd edition (1987), Eq. 61.13. More generalformulas including the excess density and viscosity have been developed by L. E. Scriven, Chem. Eng. Sci., 12,98–108 (1960).www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 2 of 12The boundary condition will be derived here by using an integral formulation. Consider twoimmiscible fluids in space that have an arbitrary two-dimensional boundary. In order for theoutward normal unit vector of this boundary to be coincident with nI , the normal unit vectorpointing into phase I, let the fluid inside the boundary be phase II and let the fluid outside bephase I.Apply Newton’s second law to the boundary.XF maSince there is no interfacial mass, the right side is zero.XF 0The collection of forces (per unit area) acting on the boundary from one phase can be obtainedfrom the second-order stress tensor π pδ τ . From it, the forces resulting from the pressure,the normal stress due to viscosity, and the shearing stress due to viscosity will be accounted for.Since there are two phases, there will be two respective tensors, π I and π II , and the total force isobtained by integrating them over the boundary S. In addition to these forces, the surfacetension σ in the boundary also needs to be considered. It’s assumed that σ is constant and thatthe tension is tangential to the boundary.This figure illustrates the forces due to surface tension in one cross-section of the boundary. As σhas units of force per unit length, the total force due to surface tension is obtained by integratingσ over the bounding curve C.www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 3 of 12Zoom in on the severed piece and draw a free-body diagram.In this figure, t is a unit vector tangential to the bounding curve C at every point. The directionof the force due to surface tension is then nI t. The subscript n is included on the stress tensorsto indicate that the forces (per unit area) are acting on a surface whose normal unit vector is nI .Newton’s second law becomes IIσ(n t) dt π · dS π II · dS 0.SCSBecause π I points towards the surface and nI points away from it, the dot product results in aminus sign. Since π II and nI both point away from the surface, this dot product is positive. 0 σ(nI t) dt π I · ( nI ) dS π II · nI dSS C S IIIII σ(n t) dt π · n dS π · nI dSS C S IIII σ(n t) dt (p δ τ ) · n dS (pII δ τ II ) · nI dSCSS IIIII σ(n t) dt [p (δ · n ) τ · n ] dS [pII (δ · nI ) τ II · nI ] dSCSSNote that Iδ·n 3 X3X δi δj δij ·i 1 j 1 3Xδj δj ·j 1 j 1 k 1www.stemjock.com!δk nIkk 1 3 X3X3X3X!δk nIkk 1δj (δj ·δk )nIk 3 X3Xj 1 k 1δj δjk nIk 3Xk 1δk nIk nI .
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 4 of 12As a result, σ(nI t) dt 0 C SI(nI pII nI · τ II ) dSS(nI pI nI · τ I nI pII nI · τ II ) dS S C[nI (pI pII ) nI · (τ I τ II )] dS.σ(nI t) dt I(n p n · τ ) dS σ(nI t) dt I S C(pII nI τ II · nI ) dS SI Iσ(n t) dt (pI nI τ I · nI ) dS SCBring the surface integral to the left side. I IIIIIIIσ(nI t) dt[n (p p ) n · (τ τ )] dS CSSince σ is constant, it can be brought in front of the line integral. I IIIIIII[n (p p ) n · (τ τ )] dS σ (nI t) dtS(1)CAll that’s left to do is to use Stokes’s theorem to turn this closed loop integral on the right sideinto a surface integral. The issue is that the integrand has a cross product rather than a dotproduct, so the theorem can’t be applied now. Consider the dot product of an arbitrary vector Awith the closed loop integral and then use the triple product vector identity,A · (B C) C · (A B). IA · (n t) dt A · (nI t) dtC C t · (A nI ) dt C (A nI ) · t dt C (A nI ) · dt(now Stokes’s law can be applied) C (A nI ) · dSS ! 3! 33XXX δi δ j Aj δk nIk · dS xiSj 1i 1k 1 ! 33 X3XX δi (δj δk )Aj nIk · dS xiSi 1j 1 k 1 ! 33 X3 X3XX δl εjkl Aj nIk · dS δi xiSj 1 k 1 l 1i 1 X3 X3 X3 X3 (δi δl )εjklAj nIk · dS xiSi 1 j 1 k 1 l 1www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 5 of 12Because the components of A are constant, they can be brought in front of the derivative. X 3 X3 X3 X3I n A · (nI t) dt (δi δl )εjkl Aj k · dS xiS i 1 j 1 k 1 l 1C 3 X3 X3 X3 X3IX n δm εilm εjkl Aj k · dS xiSi 1 j 1 k 1 l 1 m 1 3 X3 X3 X3 X3IX n δm εmil εjkl Aj k · dS xiSi 1 j 1 k 1 l 1 m 1 X3 X3 X3 X3I n δm (δmj δik δmk δij )Aj k · dS xiS i 1 j 1 k 1 m 1 3 X3 X3 X33 X3 X3 X3IIXX n n δm δmj δik Aj k δm δmk δij Aj k · dS x xiiSi 1 j 1 k 1 m 1i 1 j 1 k 1 m 1 3 X33 X3IIXX n n δ j Aj i δk Ai k · dS x xiiSi 1 j 1i 1 k 1 ! ! 3 333IXXXX n i δk nIk · dSδ j Aj Ai x xiiSi 1j 1i 1k 1 ( · nI )A A · nI · dS S [( · nI )A A · nI ] · (nI dS) S [( · nI )A · nI A · nI · nI ] dS S A · [( · nI )nI nI · nI ] dSS A·[( · nI )nI nI · nI ] dSSSimplify this second term in the integrand. ! 3!33XXX IIII n · n δiδj n j ·δk n k xii 1j 1k 1 !333 XXX nIj · δi δjδk nIk xii 1 j 1 33 X3 XXi 1 j 1 k 1www.stemjock.comk 1δi (δj · δk ) nIj In xi k
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 6 of 12Because nI is a unit vector and has a magnitude of 1, the second term in the integrand vanishes.II n · n 3 X3 X3Xδi δjki 1 j 1 k 1 3 X3Xi 1 j 13 X3X nIj In xi k nIj Iδin xi j1 (nI )22 xi ji 1 j 1 331X X I 2 δi(nj )2 xi δii 1j 13 1X δi(1)2 xii 1 0As a result, (nI t) dt A ·A· SC I [( · nI )nI n· nI ] dSnI ( · nI ) dS. A·S It then follows that InI ( · nI ) dS,(n t) dt CSand equation (1) becomes I IIIIIII[n (p p ) n · (τ τ )] dS σnI ( · nI ) dS.SSBring σ back inside the integral. [nI (pI pII ) nI · (τ I τ II )] dS nI ( · nI )σ dSSSBecause the boundary that separates the two immiscible fluids is arbitrary, the surface integralsmay be removed.nI (pI pII ) nI · (τ I τ II ) nI ( · nI )σThis equation is for the case that phase II is inside the boundary and phase I is outside of it.www.stemjock.com(2)
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 7 of 12But what if it’s the other way around?If phase I is inside and phase II is outside, then replace the I superscripts with II and replace theII superscripts with I to get the new boundary condition.nII (pII pI ) nII · (τ II τ I ) nII ( · nII )σ(3)Part (a)Here we have a spherical droplet in a medium. Let phase I be inside the droplet and let phase IIbe outside of it. Then equation (3) applies. The outward unit normal vector is r̂ if the origin ofthe coordinate system is at the sphere’s center.Set nII r̂ in equation (3).r̂(pII pI ) r̂ · (τ II τ I ) r̂( · r̂)σUse formula (A) on page 836 to calculate the divergence of r̂ in spherical coordinates (r, θ, φ).This divergence is to be evaluated at the boundary r R. 1 d 2IIIIIIσr̂(p p ) r̂ · (τ τ ) r̂ 2 (r )r drr R 2 r̂σr r R2σ r̂Rwww.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 8 of 12The second term on the left side expands as follows in spherical coordinates.IIIIIIIIIr̂ · (τ II τ I ) δr · [δr δr (τrr τrr) δr δθ (τrθ τrθ) δr δφ (τrφ τrφ)IIIIIIIII δθ δr (τθr τθr) δθ δθ (τθθ τθθ) δθ δφ (τθφ τθφ)IIIIIIIII δφ δr (τφr τφr) δφ δθ (τφθ τφθ) δφ δφ (τφφ τφφ)]IIIIIIIII (δr · δr )δr (τrr τrr) (δr · δr )δθ (τrθ τrθ) (δr · δr )δφ (τrφ τrφ)IIIIIIIII (δr · δθ )δr (τθr τθr) (δr · δθ )δθ (τθθ τθθ) (δr · δθ )δφ (τθφ τθφ)IIIIIIIII (δr · δφ )δr (τφr τφr) (δr · δφ )δθ (τφθ τφθ) (δr · δφ )δφ (τφφ τφφ)IIIIIIIII δr (τrr τrr) δθ (τrθ τrθ) δφ (τrφ τrφ)So the boundary condition becomesIIIIIIIIIr̂(pII pI ) r̂(τrr τrr) θ̂(τrθ τrθ) φ̂(τrφ τrφ) r̂2σ.RDot both sides by r̂.III(pII pI ) (τrr τrr) 2σRI τ II . Therefore, for a spherical droplet (phase I) ofSince there is no interfacial viscosity, τrrrrradius R and surface tension σ in a medium (phase II),pI pII 2σ.RR and σ are both positive, so the pressure in the droplet is less than that in the medium.pI pII 2σ 0RpI pIIwww.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 9 of 12Suppose now that there’s a flat horizontal boundary. Let phase I be on the bottom and let phaseII be on top. Assume that phase I is inside the boundary so that the outward normal unit vectoris ẑ. Then equation (3) applies once again.Substitute nII ẑ in equation (3).ẑ(pII pI ) ẑ · (τ II τ I ) ẑ( · ẑ)σUse formula (A) on page 832 to calculate the divergence in Cartesian coordinates. Thisdivergence is to be evaluated at the boundary z z0 . dIIIIIIẑ(p p ) ẑ · (τ τ ) ẑ(1)σdzz z0 0The second term on the left side expands as follows in Cartesian coordinates.IIIIIIIIIẑ · (τ II τ I ) δz · [δx δx (τxx τxx) δx δy (τxy τxy) δx δz (τxz τxz)IIIIIIIII δy δx (τyx τyx) δy δy (τyy τyy) δy δz (τyz τyz)IIIIIIIII δz δx (τzx τzx) δz δy (τzy τzy) δz δz (τzz τzz)]IIIIIIIII (δz · δx )δx (τxx τxx) (δz · δx )δy (τxy τxy) (δz · δx )δz (τxz τxz)IIIIIIIII (δz · δy )δx (τyx τyx) (δz · δy )δy (τyy τyy) (δz · δy )δz (τyz τyz)IIIIIIIII (δz · δz )δx (τzx τzx) (δz · δz )δy (τzy τzy) (δz · δz )δz (τzz τzz)IIIIIIIII δx (τzx τzx) δy (τzy τzy) δz (τzz τzz)So the boundary condition becomesIIIIIIIIIẑ(pII pI ) x̂(τzx τzx) ŷ(τzy τzy) ẑ(τzz τzz) 0.Dot both sides by ẑ.III τzz) 0(pII pI ) (τzzI τ II . Therefore, for two media separated by a planarSince there is no interfacial viscosity, τzzzzboundary,pI pII 0.The pressures are identical.pI pIIThis same result would have been obtained if phase II were assumed to be inside the boundaryinstead.www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 10 of 12Part (b)Start with Eq. 3C.5-1.IIIIIIIIIIIIII n (p p ) [n · (τ τ )] n11 R1 R2 σ(3C.5-1)Make R1 and R2 dimensionless first.IIII n (p p ) [n · (τ τ )] n1I nI nl0l0 R1 R2 1R1!σl0R2l01 σl01 R1l0 R2σl0Recall that Newton’s generalized law of viscosity gives τ as 2 †µ κ ( · v)δ.τ µ v ( v) 3For a fluid that is incompressible or of constant density, · v 0. τ µ v ( v)†The combination of velocity gradients in parentheses is a second-order tensor γ̇, which representsthe rate of fluid deformation.τ µγ̇Substitute this formula into Eq. 3C.5-1.nI (pI pII ) [nI · ( µI γ̇ I µII γ̇ II )] nI 1 R11 σl0R2 11 σI IIII IIII IIIIn (p p ) µ n · γ̇ µ n · γ̇ n l0R1 R2The units of γ̇ are the same as a velocity gradient: velocity over distance. Make γ̇ dimensionlessnow. v 0 µI Il0 Iv0 µII Il0 II11 σII IIIn (p p ) n ·γ̇ n ·γ̇ n l0v0l0v0l0R1 R2 v0 µI I I v0 µII I II11 σI IIIIn (p p ) n · γ̇ n · γ̇ n l0l0l0R1 R2 Divide both sides ρI v02 so that the right side and the last two terms on the left have familiardimensionless quantities. pIIµIµII I IIII n · γ̇ n · γ̇ nIl0 v0 ρIl0 v0 ρIρI v02IIpnwww.stemjock.com 1 R1 1 R2 σl0 v02 ρI
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 11 of 12Make it so that ρII is grouped with µII instead of ρI .nIp IIµI pIIµIIIIIII ρ n·n· nIγ̇ γ̇l0 v0 ρIl0 v0 ρIIρIρI v02I 1 R11 R2σl0 v02 ρINow make the pressure dimensionless.nII (p II p0 ) (pII p0 )µIµIIIIIII ρ n·γ̇ n·γ̇ nIl0 v0 ρIl0 v0 ρIIρIρI v02 1 R11 R2 σl0 v02 ρIInclude ρg(h h0 ) to turn these pressures in the numerator into modified pressures.nIInII[pI p0 ρI g(h h0 )] [pII p0 ρII g(h h0 )]I g(h h0 )I ρ g(h h0 ) n n22ρIρI v0v0v02 II1µIµII1σIIIIII ρn · γ̇ n · γ̇ n IIIIl0 v0 ρl0 v0 ρρl0 v02 ρIR1 R2 IIpI p0 ρI g(h h0 ) pII p0 ρII g(h h0 )I gl0 h h0I ρ gl0 h h0 n nρI v02 l0ρI v02ρI v02v02 l0 IIµII11µIσIIIII ρIn · γ̇ n · γ̇ n l0 v0 ρIl0 v0 ρIIρIl0 v02 ρIR1 R2 n (P P ) nIIIII ρII 1ρI gl0 h h0v02 l0 IIµIIµIIII ρIIγ̇ γ̇n·n· nI l0 v0 ρIl0 v0 ρIIρI n (P P ) nIIIII ρII ρIρI 1 1 σl0 v02 ρI σl0 v02 ρI R1R211gl0 hv02 IIµIIµIIIIII ρ n·γ̇ n·γ̇ nIl0 v0 ρIl0 v0 ρIIρI R1 R2Therefore, the dimensionless boundary condition is n (P P ) nIIIII ρII ρIgl0 hρIv02 I II II µµρ11σIIIIII ,[n · γ̇ ] [n · γ̇ ] I n l0 v0 ρIl0 v0 ρIIρl0 v02 ρIR1 R2where the thick square brackets enclose dimensionless combinations of variables. From page 98: 2 l0 v0 ρv0σRe Reynolds numberFr Froude numberWe Weber number.µgl0l0 v02 ρwww.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5Page 12 of 12Part (c)Example 3.7-2 considers the flow of oil (phase I) in a large tank that is exposed to the atmosphere(phase II) on top. Because the density of oil is far greater than that of air (ρI ρII ), the ratioρII /ρI is essentially zero. ρII ρIgl0 n (P P ) nhIρv02 {z }IIIII 1 II II µIσµρ11IIIIII[n·[n·γ̇] γ̇] n l0 v0 ρIl0 v0 ρIIρIl0 v02 ρI {z }R1 R2 0The gas-liquid boundary has an outward normal unit vector nI n. I gl0 11σµIn(P I P II ) n 2 h γ̇] n [n· l0 v0 ρIv0l0 v02 ρIR1 R2 Comparing the formulas for P and P II , the modified pressure for air is negligible compared toI that for oil. Not only is pI pII , but also the ratio ρII /ρI appears in P II . I 11σgl0 µI[n · γ̇ ] n nP n 2 h l0 v0 ρIv0l0 v02 ρIR1 R2 ISince the tank is large and oil is very viscous, the spinning impeller at the bottom is unlikely to make the gas-liquid boundary nonplanar. That means the sum of (1/ R1 ) and (1/ R2 ) is roughlyzero. I gl0 µII[n·γ̇] 0nP n 2 h l0 v0 ρIv0Since only quantities related to the oil are remaining, remove the I superscripts. Also, substitute h (h h0 )/l0 . gl0 h h0µnP n 2 [n · γ̇] 0l0l0 v0 ρv0 h h0 is a vertical elevation, so replace it with z. Also, use the characteristic quantities, v0 DNand l0 D. gDµznP n [n · γ̇] 0222D N DD Nρ Therefore, substituting z z/D, gµz nP n[n · γ̇] 0.22DND Nρ www.stemjock.com(3.7-36)
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3C.5 Page 3 of 12 Zoom in on the severed piece and draw a free-body diagram. In this figure, t is a unit vector tangential to the bounding curve Cat every point. The direction of the force due to surface tension is then nI t. The subscript nis included on the stress tensors
Griffiths Quantum Mechanics 3e: Problem 2.43 Page 2 of 10 The only way a function of tcan be equal to a function of xis if both are equal to a constant E. i 0(t) (t) 2 2m 00(x) (x) (x) E As a result of using the method of separation of variables, the Schr odinger equation has reduced to two ODEs, one in xand one in t. i 0(t) (t .
Griffiths Quantum Mechanics 3e: Problem 2.36 Page 3 of 6 Assume that C 1 6 0, divide both sides by C 1, and use the fact that sin2x 2sinxcosx. sin 2 p 2m a! 0 This impl
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2010 AMC 8 Problems Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6. Using only pennies, nick els, dimes, and quar ters, what is the smallest number of coins F reddie would need so he cou
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Aber es gibt natürlich auch Jungs, die beste Freunde sind, und Mädchen, die Cliquen haben. Oder Mädchen und Jungs, die befreundet sind. Doch nicht jeder findet sofort einen Freund oder eine Freundin. Dann steht man womöglich alleine herum und bekommt schnell das Gefühl, dass andere gegen einen sind. Zum Beispiel, wenn in Deiner Straße die Nachbarsmädchen über Dich lästern, weil Du ein .
