Figure 1.1: Fluid flow in a plane narrow slit.

Transcription

1 Poblem Conside a fluid of density ρ in incompessible, lamina flow in a plane naow slit of length L and width W fomed by two flat paallel walls that ae a distance 2B apat. End effects may be neglected because B W L. The fluid flows unde the influence of both a pessue diffeence p and gavity. Figue.: Fluid flow in a plane naow slit. a Using a diffeential shell momentum balance, detemine expessions fo the steady-state shea stess distibution and the velocity pofile fo a Newtonian fluid of viscosity µ. b Obtain expessions fo the maximum velocity, aveage velocity and the mass flow ate fo slit flow. Solution.a Diffeential Shell Momentum Balance in Rectangula Catesian Coodinates. A shell momentum balance is used below to deive a geneal diffeential equation that can be then employed to solve seveal fluid flow poblems in ectangula Catesian coodinates. Fo this pupose, conside an incompessible fluid in lamina flow unde the effects of both pessue and gavity in a system of length L and width W, which is at an angle β to the vetical. End effects ae neglected assuming the dimension of the system in the x-diection is elatively vey small compaed to those in the y-diection W and the z-diection L. Figue.2: Diffeential ectangula slab shell of fluid of thickness x used in z-momentum balance fo flow in ectangula Catesian coodinates. The y-axis is pointing outwad fom the plane of the compute sceen.

2 Since the fluid flow is in the z-diection, u x =, u y =, and only u z exists. Fo small flow ates, the viscous foces pevent continual acceleation of the fluid. So, u z is independent of z and it is meaningful to postulate that the velocity u z = u z x and the pessue p = pz. The only nonvanishing components of the stess tenso ae τ xz = τ zx, which depend only on x. Conside now a thin ectangula slab shell pependicula to the x-diection extending a distance W in the y-diection and a distance L in the z-diection. A ate of z-momentum balance ove this thin shell of thickness x in the fluid is of the fom: Rate of z-momentum = In Out + Geneation = Accumulation At steady-state, the accumulation tem is zeo. Momentum can go in and out of the shell by both the convective and molecula mechanisms. Since u z x is the same at both ends of the system, the convective tems cancel out because ρu 2 zw x z= = ρu 2 zw x z=l. Only the molecula tem LW τ xz emains to be consideed, whose in and out diections ae taken in the positive diection of the x-axis. Geneation of z-momentum occus by the pessue foce acting on the suface pw x and the gavity foce acting on the volume [ρg cos βlw x]. The diffeent contibutions may be listed as follows: ate of z-momentum in by viscous tansfe acoss suface at x is LW τ xz x ate of z-momentum out by viscous tansfe acoss suface at x + x is LW τ xz x + x ate of z-momentum in by oveall bulk fluid motion acoss suface at z = is ρw xu 2 zz = ate of z-momentum out by oveall bulk fluid motion acoss suface at z = L is ρw xu 2 zz = L pessue foce acting on suface at z = is p W x pessue foce acting on suface at z = L is p L W x gavity foce acting in z-diection on volume of ectangula slab is ρg cos βlw x On substituting these contibutions into the z-momentum balance, we get LW τ xz x LW τ xz x + x + p p L W x + ρg cos βlw x = Dividing equation.a. by LW x yields.a. τ xz x + x τ xz x = p p L + ρgl cos β.a.2 x L On taking the limit as x, the left-hand side of the above equation is exactly the definition of the deivative. The ight-hand side may be witten in a compact and convenient way by intoducing the modified pessue P, which is the sum of the pessue and gavitational tems. The geneal definition of the modified pessue is P = p + ρgh, whee h is the distance upwad in the diection opposed to gavity fom a efeence plane of choice. The advantages of using the modified pessue P ae that i the components of the gavity vecto g need not be calculated; ii the solution holds fo any flow oientation; and iii the fluid may flow as a esult of a pessue diffeence, gavity o both. Hee, h is negative since the z-axis points downwad, giving h = z cos β and theefoe P = pρgz cos β. Thus, P = p at z = and P L = p L ρgl cos β at z = L giving p p L + ρgl cos β = P P L P. Thus, equation.a.2 yields dτ xz dx = P L This fist-ode diffeential equation may be simply integated to give.a.3 τ xz x = P L x + C.a.4 Hee, C is an integation constant, which is detemined using an appopiate bounday condition based on the flow poblem. Equation.a.4 shows that the momentum flux o shea stess distibution is linea in systems in ectangula Catesian coodinates. Since equations.a.3 and.a.4 have been deived without making any assumption about the type of fluid, they ae applicable to both Newtonian and non-newtonian fluids. Some of the axial flow poblems in ectangula Catesian coodinates whee these equations may be used as stating points ae given below. 2

3 .b Detemination of solution. Since the poblem is a lamina flow in a plane naow slit, the obvious choice fo a coodinate system is Catesian coodinates. The fluid flow is in the z diection, theefoe the velocity vecto is of the fom v =,, v z. Futhe, v z is independent of z and it is meaningful to postulate that v z is a function only fo x, v z = v z x, and also fo the pessue p = p z. Fom these, and fom the definition of the stess tenso, its nonvanishing components ae τ xz = τ zx, which depend only on x. Using the diffeential shell momentum balance method, conside now a thin ectangula slab shell pependicula to the x-diection extending a distance W in the y-diection and a distance L in the z-diection. A ate of z-momentum balance ove this thin shell of thickness x in the fluid is of the fom: Rate of z-momentum = In Out + Geneation = Accumulation At steady-state, the accumulation tem is zeo. Momentum can go in and out of the shell by both the convective and molecula mechanisms. Since v z is a function only of x, it is the same at both ends of the slit, and the convective tems cancel out because ρv z v z W x z= = ρv z v z W x z=l. Only the molecula tem LW τ xz emains to be consideed. Geneation of z-momentum occus by the pessue foce ρw x and gavity foce ρgw L x. On substituting these contibutions into the z-momentum balance, we get LW τ xz x LW τ xz x+ x + p p L W x + ρgw L x = Dividing the above equation by LW x yields.b. τ xz x+ x τ xz x x = p p L + ρgl L.b.2 On taking the limit as x the left tem of the above equation is the definition of the deivative dτxz dx. The ight-hand side may be compactly and conveniently witten by intoducing the modified pessue P, which is the sum of the pessue and gavitational tems. The geneal definition of the modified pessue is P = p + ρgh, whee h is the distance upwad in the diection opposed to gavity fom a efeence plane of choice. Since the z-axis points downwad in this poblem, h = z and theefoe P = pρgz. Thus, P = p at z = and P L = p L ρgl at z = L, giving p p L + ρgl = P P L P. Then equation.b.2 becomes dτ xz dx = P.b.3 L Equation.b.3 on integation leads to the following expession fo the shea stess distibution tenso: τ xz = P L x + C To find the velocity, we substitute τ xz fom.b.4 into Newton s law of viscosity, and get:.b.4 µ dv z dx = P L x + C.b.5 The above fist-ode diffeential equation is simply integated to obtain the following velocity pofile: v z = P 2µL x2 C µ + C 2.b.6 It is woth noting that equations.b.3 and.b.4 apply to both Newtonian and non-newtonian fluids, and povide stating points fo many fluid flow poblems in ectangula Catesian coodinates. Using the no-slip bounday conditions at the two fixed walls, v z x x=b = v z x x= B =, the integation P B2 2µL. constants ae C = and C 2 = Then the final expesions fo the shea tenso and the velocity ae: v z = τ xz = P L x P B2 2µL x2 B 2.b.7.b.8 3

4 It is obseved that the velocity distibution fo lamina, incompessible flow of a Newtonian fluid in a plane naow slit is paabolic. Fom the velocity pofile, vaious useful quantities may be deived. The maximum velocity occus at x = whee dvz dx =, d 2 v z dx 2 <. Theefoe, v z max = v z x= = P B2 2µL The aveage velocity is obtained by dividing the volumetic flow ate by the coss-sectional aea,.b.9 v z avg = B B v zw dx B B W dx = B P B2 v z dx = 2B B 3µL = 2 3 v z max.b. Thus, the atio of the aveage velocity to the maximum velocity fo Newtonian fluid flow in a naow slit is 2 3. The mass ate of flow is obtained by integating the velocity pofile ove the coss section of the slit as follows: w = B B ρv z W dx = 2ρW Bv z avg and substituting v z avg fom.b. we have the final expesion fo the mass ate of flow..b. w = 2 P B3 ρw 3µL.b.2 The flow ate vs. pessue dop w vs. P expession above is the slit analog of the Hagen-Poiseuille equation oiginally fo cicula tubes. It is a esult woth noting because it povides the stating point fo ceeping flow in many systems e.g., adial flow between two paallel cicula disks; and flow between two stationay concentic sphees. Finally, it may be noted that the above analysis is valid when B W. If the slit thickness B is of the same ode of magnitude as the slit width W, then v z = v z x, y, i.e., v z is not a function of only x. If W = 2B, then a solution can be obtained fo flow in a squae duct. 4

5 Poblem 2 Steady, lamina flow occus in the space between two fixed paallel, cicula disks sepaated by a small gap 2b. The fluid flows adially outwad owing to a pessue diffeence P P 2 between the inne and oute adii and 2, espectively. Neglect end effects and conside the egion 2 only. Such a flow occus when a lubicant flows in cetain lubication systems. Figue 2.: Radial flow between two paallel disks. a Simplify the equation of continuity to show that u is a function of only z. b Simplify the equation of motion fo incompessible flow of a Newtonian fluid of viscosity µ and density ρ. c Obtain the velocity pofile assuming ceeping flow. d Sketch the velocity pofile u, z and the pessue pofile P. e Detemine an expession fo the mass flow ate by integating the velocity pofile. f Deive the mass flow ate expession in e using an altenative shot-cut method by adapting the plane naow slit solution. Solution 2.a Simplification of continuity equation. Since the steady lamina flow is diected adially outwad, only the adial velocity component u exists. The tangential and axial components of velocity ae zeo, so the velocity vecto becomes u = u,,. Fo an incompessible flow, the continuity equation gives u =. In cylindical coodinates, this is expanded as follows: u + u θ θ + u z z = u = u = f θ, z We ae expecting the solution to be symmetic aound the z-axis, theefoe the solution must be independent of θ, hence This poves that u is a function of z only. u = f z 2.a. 2.b Simplification of the equation of motion fo a Newtonian fluid. Fo a Newtonian fluid, its equation of motion is the Navie-Stokes equation, ρ D u Dt = P + µ 2 u 2.b. in which P includes both the pessue and gavitational tems. On noting that u = u, z, equation 2.b. can be sepaated into its coodinate components and fo steady flow they may be simplified as given below. 5

6 component: u ρ u θ component: = P θ z component: = P z = P + u µ 2 z 2 2.b.2 2.b.3 2.b.4 Fom 2.b.3 and 2.b.4 it is easy to pove that P = P. Recall that u = fz fom the continuity equation. Substituting u = f/ and P = P in equation 2.b.2 then gives ρ f 2 z dp 3 = + µ d 2 f z d dz 2 2.b.5 This is the equation of motion fo an incompessible, steady lamina flow of a Newtonian fluid of viscosity µ and density ρ. 2.c Detemination of the velocity pofile assuming ceeping flow. The equation of motion 2.b.5 is an inhomogeneous and geneally non-linea diffeential equation it depends on the fom of fz. It has no solution unless the nonlinea tem that is, the f 2 tem on the left-hand side is neglected. This is the ceeping flow assumption, and unde this assumption, equation 2.b.5 becomes an homogeneous equation. The latte can be witten as dp = µ d2 f z d dz 2 2.c. The left tem of 2.c. is a function of only, while the ight tem is a function of only z. The equality can be tue only if both tems ae equal with a constant a R, so that dp = µ d2 f z d dz 2 = a Thus the oiginal patial diffeential equation has been decoupled into two odinay diffeential equations dp d = a 2.c.2 µ d2 f z dz 2 = a 2.c.3 To solve this system, all we need to do is integate each equation equation 2.c.3 needs to be integated twice. We thus find P 2 P = a ln 2 2.c.4 f z = a 2µ z2 + a z + a 2 2.c.5 whee a and a 2 ae integation constants. By substituting a fom equation 2.c.4 into equation 2.c.5, we finally get that The no-slip bounday conditions of ou poblem dictate that f z = P 2 P z 2 + a z + a 2 2.c.6 2µ ln 2 6

7 u, z = +b = u, z = b = If we inset these into 2.c.6, the esult is that f z = P 2 P z b 2 2 2µ ln 2 b 2 theefoe as we have assumed ceeping flow, the velocity pofile is: u, z = P 2 P z b 2 2 2µ ln 2 b 2 Futhemoe the expession fo the pessue P is given fom the solution of 2.c.2 and is ln 2 P = P + [P 2 P ] ln 2 whee we have substituted a fom 2.c.4. 2.c.7 2.c.8 2.c.9 2.d Sketch of the velocity and pessue pofiles. The velocity pofile is descibed by equation 2.c.8 and the pessue pofile by equation 2.c.9. We poduce a sketch fo each using gnuplot, with the following numeical values: b =., µ = 2 = 5, = P 2 =, P = P u b.5. a Pessue pofile. b Velocity pofile. Figue 2.2: Velocity and pessue pofiles. Regadless of the pecise numeical values, the gaphs ae epesentative of the fluid s behavio when the pessue is geate at the cente. The fluid begins to flow outwads apidly, but both the velocity and pessue dop as the fluid appoaches the oute peimete. The fluid s velocity also diminishes as one appoaches eithe plate, and emains highest at the cental aea between the plates. 7

8 2.e Mass flow ate by integation of the velocity pofile. The mass flow ate w is igoously obtained by integating the velocity pofile using w = ˆn ρ uds whee ˆn is the unit nomal to the element of suface aea ds and u is the fluid velocity vecto. Fo the adial flow between paallel disks, ˆn = ˆ, u = u ˆ, and ds = 2πdz. Then, substituting the velocity pofile and integating gives w = = b b b = 2π b b ρ n u ds ρu ds b ρu, zdz and using 2.a. and 2.c.7 fo u we finally have fo the mass flow ate that w = 4π 3µ P P 2 ln 2 b 3 ρ 2.e. Mass flow ate using shot cut method by adapting naow slit solution The plane naow slit solution may be applied locally by ecognizing that at all points between the disks, the flow esembles the flow between paallel plates povided u is small that is, the ceeping flow is valid. The mass flow ate fo a Newtonian fluid in a plane naow slit of width W, length L and thickness 2B is given by w = 2 P B3 W ρ. 3µL In this expession, P/L is eplaced by dp/d, B by b, and W by 2π. Note that mass is conseved, so w is constant. Then, integating fom to 2 gives the same mass flow ate expession [equation 2.e.], as shown below. w 2 d = 4π b3 P2 dp w = 4π 3µ P 3µ P P 2 ln 2 b 3 ρ 2.e.2 This altenative shot cut method fo detemining the mass flow ate stating fom the naow slit solution is vey poweful because the appoach may be used fo non Newtonian fluids whee analytical solutions ae difficult to obtain. 8

9 Poblem 3 A wie coating die essentially consists of a cylindical wie of adius κr κ < moving hoizontally at a constant velocity u along the axis of a cylindical die of adius R. If the pessue in the die is unifom, then the polyme melt which may be consideed a non-newtonian fluid descibed by the powe law model and of constant density ρ flows though the naow annula egion solely by the dag due to the axial motion of the wie which is efeed to as axial annula Couette flow. Neglect end effects and assume an isothemal system. Figue 3.: Fluid flow in wie coating die. a Establish the expession fo the steady-state velocity pofile in the annula egion of the die. Simplify the expession fo n = Newtonian fluid. b Obtain the expession fo the mass flow ate though the annula die egion. Simplify the expession fo n = and n = /3. c Estimate the coating thickness δ some distance downsteam of the die exit. d Find the foce that must be applied pe unit length of the wie. e Wite the foce expession in d as a plane slit fomula with a cuvatue coection. Solution 3.a Shea stess and velocity distibution. The fluid flows in the z-diection sheaing constant- sufaces. Theefoe, the velocity component that exists is u z and the shea stess component is τ z. A momentum balance in cylindical coodinates gives dτ z = P d L 3.a. Since thee is no pessue gadient and gavity does not play any ole in the motion of the fluid, P =. Equation 3.a. with the ight-hand side set to zeo leads to the following expession fo the shea stess distibution on integation: τ z = C 3.a.2 The constant of integation C is detemined late using bounday conditions. Since the velocity u z deceases with inceasing adial distance, the velocity gadient du z /d is negative in the annula die egion and the powe law model may be substituted in the following fom fo τ z in equation 3.a.2. m du n z = C 3.a.3 d Hee, m and n ae the consistency index and exponent in the powe law model, espectively. The Newtonian fluid model has m = µ and n =, whee µ is the fluid s dynamic viscosity. The above diffeential equation is simply integated to obtain the following velocity pofile fo n : u z = C n m q q + C 2 3.a.4 whee q = n. The integation constants C and C 2 ae evaluated fom the following no-slip bounday conditions: 9

10 u z = R =, 3.a.5 u z = κr = V 3.a.6 Fom 3.a.5, So, and 3.a.6 yields u z = C 2 = C n m C n m R q q R q q [ q ] R V = C n m R q q κq On substituting the integation constants into equations 3.a.2 and 3.a.4, the final expessions fo the shea stess and velocity distibution fo n ae τ z = m [ ] n qv R q κ q 3.a.7 u z = /Rq V κ q 3.a.8 When n =, the ight-hand sides of equations 3.a.7 and 3.a.8 evaluate to /; theefoe, the evaluation may be done using De l Hopital s ule by diffeentiating the numeato and denominato with espect to n o q and then evaluating the limit as n o q. On diffeentiating equation 3.a.7, lim q q κ q Similaly, on diffeentiating equation 3.a.8, [ /R q ] lim q κ q / = lim q / = lim q κ q ln κ = ln κ [ /R q ] ln /R κ q = ln κ ln /R ln κ Thus, on simplifying equations 3.a.7 and 3.a.8, the shea stess and velocity fo a Newtonian fluid n = and m = µ ae τ z = u z V = µv ln /κ ln /R ln κ 3.a.9 3.a. 3.b Mass flow ate. The mass ate of flow is obtained by integating the velocity pofile ove the coss section of the annula egion as follows. Integation then gives w = 2πρ R κr u z d = 2πR2 V ρ κ q κ [ q ] d R R R

11 is [ w = 2πR2 V ρ 2+q ] 2 κ q 2 + q R 2 R κ 3.b. On evaluation at the limits of integation, the final expession fo the mass ate of flow fo n and n 3 w = 2πR2 V ρ κ q [ κ 2+q 2 + q ] κ2 2 3.b.2 When n = o q =, the ight-hand side of the above equation gives /. Again, using De l Hopital s ule gives [ κ 2+q lim q κ q 2 + q ] κ2 2 [ / = lim q κ q ln κ = 2κ2 ln κ + κ 2 4 ln κ 2 + q κ 2+q ln κ κ 2+q ] 2 + q 2 Substituting this in equation 3.b.2 gives the mass flow ate fo a Newtonian fluid n = as [ ] w = πr2 V ρ κ 2 2 ln /κ 2κ2 When n = /3 o q = 2, the tem κ 2+q /2 + q becomes /. Again using l Hopital s ule, κ 2+q / lim = lim κ 2+q ln κ = ln κ q q q 2 3.b.3 Substituting this last esult in equation 3.b.2 gives the mass flow ate fo a fluid with n = /3 q = 2 as ] w = 2πR2 V ρ κ κ2 [ln κ q 3.b c Coating thickness. Some distance downsteam of the die exit, the liquid polyme melt is tanspoted as a coating of thickness δ exposed to ai. A momentum balance in the ai egion yields the same shea stess expession [equation 3.a.2]. The bounday condition at the liquid-gas inteface is τ z = at = κr + δ, which now gives C = and τ z = thoughout the coating. Futhemoe, equations 3.a.4 and 3.a.6 ae valid and give the velocity pofile as u z = V though the entie coating. As the velocity becomes unifom and identical to that of the wie, the liquid is tanspoted as a igid coating. The mass flow ate in the ai egion is given by [ κr+δ w ai = 2πρV d = πr 2 V ρ κ + δ ] 2 κ 2 3.c. R κr Equating the above expession fo the flow ate w ai in the ai egion to the flow ate w in the die egion gives δ R = κ 2 + w πr 2 V ρ κ 3.c.2 On substituting the appopiate expession fo w fom equations 3.b.4, 3.b.3, o 3.b.2, the final expession fo the coating thickness is obtained. It may be noted that the coating thickness δ depends only on R, κ and n, but not on V and m.

12 3.d Foce applied to the wie. The foce applied to the wie is given by the shea stess integated ove the wetted suface aea. Theefoe, on using equation 3.a.7 o equation 3.a.9, { n F z L = 2πκR [τ z] =κr F qv z L = 2πm R q κ, n q 2πµV ln /κ, n = 3.d. 3.e Plane slit fomula fo foce. A vey naow annula egion with oute adius R and inne adius ɛr, whee ɛ is vey small, may be appoximated by a plane slit with aea = 2π ɛ 2 RL 2πRL and gap = ɛr. Then, the velocity pofile in the plane slit will be linea and the foce applied to the plane wie is given by F plane 2πRL = µ V ɛr o F plane = 2πµV 3.e. L ɛ On substituting κ = ɛ since R κr = ɛr fo the thin annula gap in equation 3.d., the foce applied to the wie is given by F z L = 2πµV ln ɛ The Taylo seies expansion given below is next used. ln ɛ = ɛ 2 ɛ2 3 ɛ3 4 ɛ4 5 ɛ e.2 3.e.3 Theefoe, the denominato of equation 3.e.2 may be witten as ln ɛ = ɛ + 2 ɛ + 3 ɛ2 + 4 ɛ3 + 5 ɛ = ɛ 2 ɛ 2 ɛ2 24 ɛ ɛ e.4 Equation 3.e.4 may be obtained by using + u = u + u 2 u 3 + u 4... which is an infinite geometic pogession with u <. On substituting in equation 3.e.2, the following foce expession is finally obtained as a plane slit fomula with a cuvatue coection. F z L = 2πµV ɛ [ 2 ɛ 2 ɛ2 24 ɛ3 9 ] 72 ɛ e.5 The above poblem solution consides only dag flow in a wie-coating die. The poblem is solved consideing both dag and pessue by Malik, R. and Shenoy, U. V. 99 Genealized annula Couette flow of a powe-law fluid, Ind. Eng. Chem. Res. 3 8:

13 Poblem 4 Conside an incompessible isothemal fluid in lamina flow between two coaxial cylindes, whose inne and oute wetted sufaces have adii of κr and R, espectively. The inne and oute cylindes ae otating at angula velocities Ω i and Ω o, espectively. End effects may be neglected. Figue 4.: Tangential annula flow between two slowly otating cylindes. a Detemine the steady-state velocity distibution in the fluid fo small values of Ω i and Ω o. b Find the toques acting on the two cylindes duing the tangential annula flow of a Newtonian fluid. Solution 4.a Steady-state velocity distibution. Simplification of the continuity equation In steady lamina flow, the fluid is expected to tavel in a cicula motion fo low values of Ω i and Ω o. Only the tangential component of velocity exists. The adial and axial components of velocity ae zeo; so, u = and u z =. Fo an incompessible fluid, the continuity equation gives u =. In cylindical coodinates, u + u θ θ + u z z = u θ θ = So, u θ = u θ, z. If end effects ae neglected, then u θ does not depend on z. Thus, u θ = u θ. Simplification of the equation of motion Thee is no pessue gadient in the θ-diection. simplify to Theefoe, the components of the equation of motion component: ρ u2 θ = p θ component: = d d [ d d u θ z component: = p z ρg ] z points upwads whee in the equation fo the θ-component the patial deivatives have been eplaced with odinay ones, since u θ is a function only of as was shown in the pevious step. The -component povides the adial pessue distibution due to centifugal foces and the z-component gives the axial pessue distibution due to gavitational foces the hydostatic head effect. Solution of the diffeential equation and the velocity pofile The θ-component of the equation of motion can be integated to get the velocity pofile: 3

14 d d u 2 θ = C u θ = C 2 + C 2 u θ = C 2 + C 2 The no-slip bounday conditions at the two cylindical sufaces ae 4.a. u θ = κr = Ω i κr Ω i κr = C 2 κr + C 2 κr u θ = R = Ω o R Ω o R = C 2 R + C 2 R 4.a.2 4.a.3 Solving equations 4.a.2 and 4.a.3 fo the integation constants we find that they ae given by Ω i Ω o κr = C 2 κr κ, R R Ω i κr Ω o = C κr R κ 2 κ Substituting the above values fo C and C 2 in equation 4.a., the velocity pofile is obtained as u θ = Ω o Ω i κ 2 κ 2 + Ω i Ω o κ 2 R 2 κ 2 = κr [ κ 2 Ω o Ω i κ 2 κr + Ω i Ω o κr ] 4.a.4 The velocity distibution given by the above expession can be witten in the following altenative fom: u θ = Ω oκr κ 2 κr κr + Ω iκ 2 R R κ 2 4.a.5 R In the above fom, the fist tem on the ight-hand-side coesponds to the velocity pofile when the inne cylinde is held stationay and the oute cylinde is otating with an angula velocity Ω o. This is essentially the configuation of the Couette - Hatschek viscomete fo detemining viscosity by measuing the ate of otation of the oute cylinde unde the application of a given toque as discussed below. The second tem on the ight-hand-side of the above equation coesponds to the velocity pofile when the oute cylinde is fixed and the inne cylinde is otating with an angula velocity Ω i. This is essentially the configuation of the Stome viscomete fo detemining viscosity by measuing the ate of otation of the inne cylinde unde the application of a given toque as discussed next. 4.b Toques acting on the cylindes. Detemination of momentum flux Fom the velocity pofile, the momentum flux shea stess is detemined as τ θ = µ d d uθ = 2µ Ω i Ω o κ 2 R 2 κ b. Detemination of toque The toque is obtained as the poduct of the foce and the leve am. The foce acting on the inne cylinde itself is the poduct of the inwad momentum flux and the wetted suface aea of the inne cylinde. Thus, T z = 2πκRLκR τ θ =R = 4πµΩ i Ω o κ 2 R2 L 4.b.2 The tangential annula flow system above povides the basic model fo otational viscometes to detemine fluid viscosities fom measuements of toque and angula velocities and fo some kinds of fiction beaings. κ 2 4

15 Poblem 5 An incompessible isothemal liquid in lamina flow is open to the atmosphee at the top. Detemine the shape of the fee liquid suface z at steady state neglecting end effects, if any fo the following cases: a case. b case. c case. d case. Figue 5.: Fee suface shape of liquid in fou cases of tangential flow. a when the liquid is in the annula space between two vetical coaxial cylindes, whose inne and oute wetted sufaces have adii of κr and R, espectively. Hee, the inne cylinde is otating at a constant angula velocity Ω i and the oute cylinde is stationay. Let z R be the liquid height at the oute cylinde. b when a single vetical cylindical od of adius R i is otating at a constant angula velocity Ω i in a lage body of quiescent liquid. Let z Ro be the liquid height fa away fom the otating od. Hint: Use the esult fom a. c when the liquid is in the annula space between two vetical coaxial cylindes, whose inne and oute wetted sufaces have adii of κr and R, espectively. Hee, the oute cylinde is otating at a constant angula velocity Ω o and the inne cylinde is stationay. Let z R be the liquid height at the oute cylinde. d when the liquid is in a vetical cylindical vessel of adius R, which is otating about its own axis at a constant angula velocity Ω o. Let z R be the liquid height at the vessel wall. Hint: Use the esult fom c. Solution Simplification of the continuity equation In steady lamina flow, the liquid is expected to tavel in a cicula motion with only the tangential component of velocity. The adial and axial components of velocity ae zeo; so, u = and u z =. Fo an incompessible fluid, the continuity equation gives u =. In cylindical coodinates, u + u θ θ + u z z = u θ θ = 5. So, u θ = u θ, z. If end effects ae neglected, then u θ does not depend on z. Thus, u θ = u θ. 5

16 Simplification of the equation of motion Thee is no pessue gadient in the θ-diection. The components of the equation of motion simplify to component: ρ u2 θ = p θ component: = d d [ d d u θ z component: = p z ρg ] z points upwads whee in the equation fo the θ-component the patial deivatives have been eplaced with odinay ones, since u θ is a function only of as was shown in the pevious step. As discussed late, the -component povides the adial pessue distibution due to centifugal foces and the z-component gives the axial pessue distibution due to gavitational foces the hydostatic head effect. Solution of the diffeential equation and the velocity pofile The θ-component of the equation of motion is integated to get the velocity pofile: [ ] d d d d u θ = d d u 2 θ = C u θ = C 2 + C and theefoe, the velocity pofile is descibed by the equation u θ = C 2 + C 2 The no-slip bounday conditions at the two cylindical sufaces ae 5.3 u θ = κr = Ω i κr Ω i κr = C 2 κr + C 2 κr u θ = R = Ω o R Ω o R = C 2 R + C 2 R Solving equations 5.4 and 5.5 fo the integation constants we find that they ae given by Ω i Ω o κr = C 2 κr κ, R R Ω i κr Ω o = C κr R κ 2 κ Substituting the above values fo C and C 2 in equation 5.3, the velocity pofile is obtained as u θ = Ω o Ω i κ 2 κ 2 + Ω i Ω o κ 2 R 2 κ 2 = κr [ κ 2 Ω o Ω i κ 2 κr + Ω i Ω o κr ] The velocity distibution given by the above expession can be witten in the following altenative fom: u θ = Ω oκr κ 2 κr κr + Ω iκ 2 R R κ R In the above fom, the fist tem on the ight-hand-side coesponds to the velocity pofile when the inne cylinde is held stationay and the oute cylinde is otating with an angula velocity Ω o. The second tem on the ight-hand-side of the above equation coesponds to the velocity pofile when the oute cylinde is fixed and the inne cylinde is otating with an angula velocity Ω i

17 5.a Fee suface shape fo otating inne cylinde and fixed oute cylinde. Now, conside the case whee the inne cylinde is otating at a constant angula velocity Ω i and the oute cylinde is stationay. With Ω o =, the velocity pofile in 5.7 is substituted into the -component of the equation of motion to get p = ρu2 θ = ρ Ω 2 i κ4 R 2 R 2 κ R 2 Integating the above equation we find the pessue pofile as p, z = ρω2 i κ4 R 2 κ 2 2 R2 5.a. 2 2 ln R 2 + f z 5.a.2 It is clea that the pessue pofile is independent of θ due to the axial symmety pesent. The above expession fo the pessue p is next substituted into the z-component of the equation of motion to obtain p z = ρg df dz = ρg f z = ρgz + C 3 5.a.3 On substituting the expession fo f in 5.a.3 we find that p, z = ρω2 i κ4 R 2 κ 2 2 R2 2 2 ln R 2 ρgz + C 3 The constant C 3 is detemined fom the equiement that at the oute cylinde the liquid height be z R. At that point, the pessue is equal to the atmospheic pessue, p atm, theefoe p = R, z = z R = p atm p atm = ρω2 i κ4 R 2 κ ln R ρgz R + C 3 Subtaction of the above two equations allows fo elimination of C 3 and the detemination of the pessue distibution in the liquid as p, z p atm = ρω2 i κ4 R 2 κ 2 2 R2 2 2 ln R 2 + ρgz R z 5.a.4 Since p = p atm at evey point acoss the liquid-ai inteface bounday, the shape of the fee liquid suface is obtained as z R z = 2g Ωi κ 2 R κ 2 2 [ R ] ln. 5.a.5 R R 5.b Fee suface shape fo otating cylinde in an infinite expanse of liquid. The esult in a may be used to deive the fee suface shape when a single vetical cylindical od of adius R i is otating at a constant angula velocity Ω i in a lage body of quiescent liquid. On substituting R = R o and κ = R i /R o, equations 5.7 with Ω o = and 5.a.5 yield Ω i R 2 i u θ = Ri 2/R2 o Ro 2 and z Ro z = Ωi R 2 2 [ 2 i 2g Ri /R2 o R 2 o ] ln 2 R o Ro 4 5.b. The above equations give the velocity pofile and the shape of the fee liquid suface fo the case whee the inne cylinde of adius R i is otating at a constant angula velocity Ω i and the oute cylinde of adius R o is stationay. Hee, z Ro is the liquid height fa away fom the otating od. In the limit as R o tends to infinity, the velocity pofile and fee suface shape fo a single vetical od otating in a lage body of quiescent liquid ae obtained as u θ = Ω ir 2 i and z Ro z = Ω2 i R4 i 2g 2. 5.b.2 7

18 5.c Fee suface shape fo otating oute cylinde and fixed inne cylinde. Next, conside the case whee the oute cylinde is otating at a constant angula velocity Ω o and the inne cylinde is stationay. With Ω i =, the velocity pofile in equation 5.7 is substituted into the -component of the equation of motion to get p = ρu2 θ Ω 2 oκ 2 R 2 κ 2 R 2 κ κ 2 R 2 = ρ 5.c. Integation gives p, z = ρω2 oκ 2 R 2 κ2 R 2 κ ln + 2 2κ 2 R 2 + f 2 z 5.c.2 The above expession fo the pessue p is now substituted into the z-component of the equation of motion to obtain p z = ρg df 2 dz = ρg f 2z = ρgz + C 4 5.c.3 On substituting the expession fo f 2 in equation 5.c.2, we get p, z = ρω2 oκ 2 R 2 κ2 R 2 κ ln + 2 2κ 2 R 2 ρgz + C 4 Again as in a, the constant C 4 is detemined fom the equiement that at the oute cylinde the liquid height be z R. At that point, the pessue is equal to the atmospheic pessue, p atm, theefoe p = R, z = z R = p atm p atm = ρω2 oκ 2 R 2 κ2 κ ln R + 2κ 2 ρgz R + C 4 Subtaction of the above two equations allows fo elimination of C 4 and the detemination of the pessue distibution in the liquid as p, z p atm = ρω2 oκ 2 R 2 [ κ2 R 2 κ ln ] R 2κ 2 2 R 2 + ρgz R z 5.c.4 Since p = p atm at evey point acoss the liquid-ai inteface bounday, the shape of the fee liquid suface is obtained as z R z = 2 Ωo R R [κ 4 2 ] 2g κ κ 2 ln + 2 R R 2. 5.c.5 An altenative appoach is to obtain the above equation fom the esult of a by substituting = κr in equation 5.a.5 to get z R z κr = 2g Ωi κ 2 2 R κ ln κ κ2 κ2 5.c.6 The above expession gives the diffeence in elevations of the liquid suface at the oute and inne cylindes. Subtacting equation 5.c.6 fom equation 5.a.5 yields z κr z = 2g Ωi κ 2 2 [ R R 2 κ 2 2 ] 2 κ ln κr R 2 κ2 5.c.7 The cylindes must now be swapped, i.e., the otating cylinde should be made the oute cylinde athe than the inne one. So, let κr = R o and R = βr o. On substituting κ = /β and R = βr o in the above equation, we get z Ro z = 2g 2 [ Ωi βr o β 2 R 2 o β 2 2 β ln R o β 2 Ro 2 ] β 2 5.c.8 On ecognizing that β which is a dummy paamete can be simply eplaced by κ, R o may be eplaced by R, and Ω i by Ω o, the above equation is identical to equation 5.c.5. 8

19 5.d Fee suface shape fo cylindical vessel otating about its own axis. The esult in c may be used to deive the fee suface shape when a liquid is in a vetical cylindical vessel of adius R, which is otating about its own axis at a constant angula velocity Ω o. In the limit as κ tends to zeo, equation 5.7 with Ω i = and equation 5.c.5 yield the velocity pofile and fee suface shape as u θ = Ω o R and z R z = Ω2 or 2 [ ] 2 5.d. 2g R Note that as κ tends to zeo, the fist and second tems in squae backets in equation 5.c.5 also tend to zeo. The velocity pofile suggests that each element of the liquid in a otating cylindical vessel moves like an element of a igid body. The fee suface of the otating liquid in a cylindical vessel is a paaboloid of evolution [since equation 5.d. coesponds to a paabola]. 9

20 Poblem 6 A plastic esin is in a vetical cylindical vessel of adius R, which is otating about its own axis at a constant angula velocity Ω. Figue 6.: Paabolic mio fom fee suface shape of otating liquid. a Detemine the shape of the fee suface z at steady state neglecting end effects, if any. Let z R be the liquid height at the vessel wall. b Show that the adius of cuvatue of the fee suface shape is given by [ ]3 2 / 2 dz d 2 z c = + d d 2 c The cylinde is otated until the esin hadens in ode to fabicate the backing fo a paabolic mio. Detemine the angula velocity in pm necessay fo a mio of focal length 75 cm. Hint: The focal length f is one-half the adius of cuvatue c at the axis. Solution Simplification of the continuity equation In steady lamina flow, the liquid is expected to tavel in a cicula motion with only the tangential component of velocity. The adial and axial components of velocity ae zeo; so, u = and u z =. Fo an incompessible fluid, the continuity equation gives u =. In cylindical coodinates this is witten as follows: u + u θ θ + u z z = u θ θ = So, u θ = u θ, z. If end effects ae neglected, then u θ cannot depend on z. Thus, u θ = u θ. Simplification of the equation of motion Thee is no pessue gadient in the θ-diection. The components of the equation of motion simplify to component: ρ u2 θ = p θ component: = d d [ d d u θ z component: = p z ρg ] z points upwads whee in the equation fo the θ-component the patial deivatives have been eplaced with odinay ones, since u θ is a function only of as was shown in the pevious step. The -component povides the adial pessue distibution due to centifugal foces and the z-component gives the axial pessue distibution due to gavitational foces the hydostatic head effect. 2

21 Solution of the diffeential equation and the velocity pofile The θ-component of the equation of motion can be integated to get the velocity pofile: d d u 2 θ = C u θ = C 2 + C 2 u θ = C 2 + C 2 Since the velocity u θ is finite at =, the integation constant C 2 must be zeo. Because u θ = R = ΩR, C = 2Ω. Thus, 6. u θ = Ω 6.2 The above velocity pofile suggests that each element of the liquid in a otating cylindical vessel moves like an element of a igid body. 6.a Fee suface shape fo otating liquid in cylinde. Next, the velocity pofile is substituted into the -component of the equation of motion to get p = ρu2 θ Equation 6.a. is easily integated to obtain = ρω 2 6.a. p, z = ρω f z 6.a.2 The above expession fo the pessue p is now substituted into the z-component of the equation of motion to obtain dp dz = ρg df dz = ρg f z = ρgz + C 3 6.a.3 On substituting the expession fo f in equation 6.a.2, p, z = ρω2 2 2 ρgz + C 3 The constant C 3 can be found fom the following bounday condition: p = R, z = z R = p atm p atm = ρω2 2 R2 ρgz R + C 3 whee z R is the liquid height at the vessel wall and p atm denotes the atmospheic pessue. Subtaction of the above two equations eliminates C 3 and gives the pessue distibution in the liquid as p, z p atm = ρω2 2 2 R 2 + ρgz R z 6.a.4 Since p = p atm at the liquid-ai inteface, the shape of the fee liquid suface is finally obtained as z R z = Ω2 2g R2 2 6.a.5 The fee suface of the otating liquid in a cylindical vessel is a paaboloid of evolution [since equation 6.a.5 coesponds to a paabola]. 2

22 6.b Geneal expession fo adius of cuvatue. The cicle that is tangent to a given cuve at point P, whose cente lies on the concave side of the cuve and which has the same cuvatue as the cuve has at P, is efeed to as the cicle of cuvatue. Its adius CP in Figue 6. defines the adius of cuvatue at P. The cicle of cuvatue has its fist and second deivatives espectively equal to the fist and second deivatives of the cuve itself at P. If the coodinates of the cente C and point P ae, z and, z espectively, then the adius of cuvatue CP is given by c = 2 + z z 2 6.b. Now, the adius CP is pependicula to the tangent to the cuve at P. So, the slope of the adius CP is given by d dz = z z Thus, the fist and second deivatives of the cuve at point P ae dz d = z z and d 2 z d 2 = z z + dz z z 2 d = + dz/d2 z z On substituting the above deivatives in equation 6.b., the geneal expession fo the adius of cuvatue is obtained as [ ]3 2 / 2 dz d 2 z c = + d d 2 6.c Angula velocity necessay fo paabolic mio fabication.. 6.b.2 Let z be the height of the liquid suface at the centeline of the cylindical vessel. Substituting z = z at = in equation 6.a.5 gives z R z = Ω2 R 2 2g. Thus, an altenative expession in tems of z fo the shape of the fee suface is Diffeentiating the above expession twice gives z = z + Ω2 2g 2 6.c. dz d = Ω2 g and d 2 z d 2 = Ω2 g We can use the above deivatives to calculate the adius of cuvatue of the paabola. By substituting the deivatives into equation 6.b.2, the adius of cuvatue of the paabola is given by c = + Ω4 2 3 / 2 g g 2 Ω 2 Since the focal length f of the mio is one half the adius of cuvatue c at the axis, 6.c.2 f = 2 c = = g 2Ω 2 and theefoe, the otational speed necessay to manufactue a paabolic mio of focal length f is g Ω = 2f 6.c.3 On substituting the focal length f =.75 m and the gavitational acceleation g = 9.8 m/s 2 in the above equation, the otational speed may be calculated as Ω =.68 ad/s = π pm = 6. pm. 22

23 Poblem 7 Conside an incompessible isothemal fluid in lamina flow between two concentic sphees, whose inne and oute wetted sufaces have adii of κr and R, espectively. The inne and oute sphees ae otating at constant angula velocities Ω i and Ω o, espectively. The sphees otate slowly enough that the ceeping flow assumption is valid. Figue 7.: Flow between two slowly otating sphees. a Detemine the steady-state velocity distibution in the fluid fo small values of Ω i and Ω o. b Find the toques on the two sphees equied to maintain the flow of a Newtonian fluid. c Simplify the velocity and toque expessions fo the case of a single solid sphee of adius R i otating slowly at a constant angula velocity Ω i in a vey lage body of quiescent fluid. Solution 7.a The steady-state velocity distibution. Simplification of the continuity equation In steady lamina flow, the liquid is expected to tavel in a cicula motion fo low values of Ω i and Ω o with only the tangential component of velocity. The adial and pola components of velocity ae zeo; so, u = and u θ =. Fo an incompessible fluid, the continuity equation gives u =. In spheical coodinates this is witten as follows: So, u φ = u φ, θ. 2 2 u + ηµ θ θ u θ ηµ θ + ηµ θ φ u φ = Simplification of the equation of motion When the flow is vey slow, the tem ρ u u in the equation of motion can be neglected because it is quadatic in the velocity. This is efeed to as the ceeping flow assumption in othe wods, we ignoe convective acceleation. Thus, fo steady ceeping flow, the entie left-hand side of the equation of motion is zeo and we get = p + µ 2 u + ρg = P + µ 2 u whee P is the modified pessue given by P = p + ρgh and h is the elevation in the gavitational field i.e., the distance upwad in the diection opposite to gavity fom some chosen efeence plane. In this poblem, h = z = συν θ. The above equation is efeed to as the Stokes flow equation o the ceeping flow equation of motion. The components of the ceeping flow equation in spheical coodinates simplify to 23

24 component: = P θ component: = dp dθ φ component: = 2 2 u φ + [ 2 θ ηµ θ ] u φ ηµ θ θ Thee is no dependence on the angle φ due to symmety about the vetical axis. The above patial diffeential equation must be solved fo the velocity distibution u φ, θ. Solution of the diffeential equation and the velocity pofile The no-slip bounday conditions at the two spheical sufaces ae u φ = κr, θ = Ω i κr ηµ θ 7.a. u φ = R, θ = Ω o R ηµ θ 7.a.2 Fom the bounday conditions, the fom u φ, θ = f ηµ θ appeas to be an educated guess to solve the patial diffeential equation fo the velocity pofile. On substituting this fom in the φ-component of the equation of motion and simplifying, we get 2 d2 f df + 2 d2 d 2f = 7.a.3 The above odinay diffeential equation may be solved by substituting f = n i.e. using the powe seies method. The esulting chaacteistic equation is n 2 + n 2 =, whose solution is n =, 2. Thus, f = C + C 2 2 and u φ, θ = C + C 2 2 On imposing the bounday conditions 7.a. and 7.a.2, we find that Ω i κr = C κr + C 2 κ 2 R 2 and Ω o R = C R + C 2 R 2 The integation constants ae thus given by C = Ω o Ω i κ 3 κ 3 and C 2 = Ω i Ω o κ3 R 3 κ 3 On substituting fo C and C 2 in 7.a.4, the velocity pofile is obtained as ηµ θ 7.a.4 [ Ωo Ω i κ 3 u φ, θ = κ 3 + Ω i Ω o κ3 R 3 ] [ κ 3 2 ηµ θ = κr ] 2 κr κ 3 Ω o Ω i κ 3 + Ω i Ω o ηµ θ κr 7.a.5 The velocity distibution given by the above expession can be witten in the following altenative fom: u φ, θ = [ Ωo κr κ 3 κr κ2 R Ω i κ 3 R κ 3 R 2 2 R ] ηµ θ 7.a.6 In the above fom, the fist tem on the ight-hand-side coesponds to the velocity pofile when the inne sphee is held stationay and the oute sphee is otating with an angula velocity Ω o. The second tem on the ight-hand-side of the above equation coesponds to the velocity pofile when the oute sphee is fixed and the inne sphee is otating with an angula velocity Ω i. 24

25 7.b Detemination of momentum flux and toque. Fom the velocity pofile, the momentum flux shea stess is detemined as τ φ = µ uφ = 3µ Ω i Ω o κ 3 R 3 κ 3 3 ηµ θ 7.b. The toque equied to otate the inne sphee is obtained as the integal ove the sphee suface of the poduct of the foce exeted on the fluid by the solid suface element and the leve am κr ηµ θ. The foce equied on the inne sphee is the poduct of the outwad momentum flux and the wetted suface aea of the inne sphee. Thus, the diffeential toque on the inne sphee suface element is dt z = τ φ =κr 2πκ 2 R 2 ηµ θdθκr ηµ θ = 3µ Ω i Ω o κ 3 ηµ θ2πκ2 R 2 ηµ θdθκr ηµ θ 7.b.2 On integating fom to π, the toque needed on the inne sphee is T z = 6πµΩ i Ω o κ3 R 3 κ 3 π ηµ 3 θdθ = 8πµΩ i Ω o κ3 R 3 κ 3 7.b.3 Note that the integal above evaluates to 4/3. Similaly, the diffeential toque needed on the oute sphee suface element is obtained as the poduct of the inwad momentum flux, the wetted suface aea of the oute sphee, and the leve am. Thus, dt z = τ φ =R 2πR 2 ηµ θdθr ηµ θ = 3µ Ω i Ω o κ 3 ηµ θ2πr2 ηµ θdθr ηµ θ 7.b.4 On integating fom to π, the toque needed on the inne sphee is T z = 6πµΩ o Ω i κ3 R 3 κ 3 7.c Limiting case of single otating sphee. π ηµ 3 θdθ = 8πµΩ o Ω i κ3 R 3 κ 3. 7.b.5 Fo the case of a single solid sphee of adius R i otating slowly at a constant angula velocity Ω i in a vey lage body of quiescent fluid, let R = R o and κ = R i /R o. Then, on letting Ω o =, equations 7.a.6 and 7.b.3 give R 3 u φ, θ = Ω i i R i /R o 3 2 R3 i Ro 3 ηµ θ and T z = 8πµΩ i /R i 3 /R o 3 7.c. The above equations hold when the oute sphee of adius R o is stationay and the inne sphee of adius R i is otating at an angula velocity Ω i. In the limit as R o tends to infinity, the esults fo a single otating sphee in an infinite body of quiescent fluid ae obtained as u φ, θ = Ω i R 3 i 2 ηµ θ and T z = 8πµΩ i R 3 i. 7.c.2 25

26 Poblem 8 An isothemal, incompessible fluid of density ρ flows adially outwad owing to a pessue diffeence between two fixed poous, concentic spheical shells of adii κr and R. Note that the velocity is not zeo at the solid sufaces. Assume negligible end effects and steady lamina flow in the egion κr R. Figue 8.: Radial flow between two poous concentic sphees. a Simplify the equation of continuity to show that 2 u = constant. b Simplify the equation of motion fo a Newtonian fluid of viscosity µ. c Obtain the pessue pofile P in tems of P R and u R, the pessue and velocity at the sphee of adius R. d Detemine the nonzeo components of the viscous stess tenso fo the Newtonian case. Solution 8.a Simplification of continuity equation. Since the steady lamina flow is diected adially outwad, only the adial velocity component v exists. The othe two components of velocity ae zeo; so, u θ = and u φ =. Fo an incompessible flow, the continuity equation gives u =. In spheical coodinates, this is witten in fully expanded fom as follows: 2 2 u + ηµ θ θ u θ ηµ θ + ηµ θ φ u φ = 2 u = Afte integating the above simplified continuity equation, we find that 2 u = fθ, φ. Howeve, due to the symmety pesent in the poblem, thee is no dependence expected on the angles θ and φ. In othe wods, 2 u must be equal to a constant, C. This is simply explained fom the fact that mass o volume, if density ρ is constant is conseved; so, ρ4π 2 u = w is constant, and C = w 4πρ. 8.b Simplification of the equation of motion fo a Newtonian fluid. The equation of motion is ρ D u Dt = p τ ab + ρ g 8.b. The above equation is simply Newton s second law of motion fo a fluid element. It states that, on a pe unit volume basis, the mass multiplied by the acceleation is because of thee foces, namely the pessue foce, the viscous foce, and the gavity foce. Fo an incompessible, Newtonian fluid, the tem τ ab = µ 2 u and the equation of motion yields the Navie - Stokes equation. On noting that 2 u = constant fom the continuity equation, the components of the equation of motion fo steady flow in spheical coodinates may be simplified as given below. 26

27 u component: = ρ u = p ρg συν θ θ component: = dp + ρg ηµ θ dθ φ component: = p φ Note that the modified pessue P may be defined as P = p + ρgh, whee h is the elevation o the height above some abitay datum plane, to avoid calculating the components of the gavitational acceleation vecto g in spheical coodinates. Hee, giving P = p + ρg συν θ, P = p + ρg συν θ and P θ = p ρg ηµ θ. θ Then, the components of the equation of motion in tems of the modified pessue ae u component: = ρ u = P θ component: = dp dθ φ component: = P φ 8.b.2 8.b.3 8.b.4 Note that equations 8.b.2 8.b.4 could have been diectly obtained fom the following fom of the equation of motion: ρ D u Dt = P τ ab 8.b.5 in which the modified pessue P includes both the pessue and gavitational tems. Fom equations 8.b.3 and 8.b.4, P s patial deivatives with espect to θ and φ vanish, theefoe P is a function of only. Substituting P = P and u = u in equation 8.b.2 then gives dp d = ρu du d. 8.b.6 8.c Pessue pofile. To calculate the pessue pofile, we stat by substituting u = C/ 2 into equation 8.b.6, which then gives Integation then poduces dp d = 2ρC2 5 P = ρc C 8.c. To detemine the integation constants, C and C, we make use of the bounday conditions. Since the pessue is P = R = P R and the velocity is u = R = v R, the integation constants may be evaluated as C = P R + ρ C2 2R 4 and C = R 2 u R 27

28 Then, substitution in equation 8.c. yields the pessue pofile as [ ] [ 4 P = P R + ρc2 R 2R 4 = P R + 2 ρu2 R ] 4 R 8.c.2 8.d Detemination of nonzeo components of the viscous stess tenso. Fo an incompessible, Newtonian fluid, the viscous stess tenso is given by τ ab = µ b u a + a u b. Since the only velocity component that exists is u = C/ 2, the nonzeo components of the viscous stess tenso ae τ = 2µ du d = 4µC 3 and τ θθ = τ φφ = 2µ u = 2µC 3 8.d. Note that the shea stesses ae all zeo. The nomal stesses in equation 8.b.6 ae patly compessive and patly tensile. Not only is the φ-component of τ diectly zeo, but the -component and the θ-component ae zeo too as evaluated below. [ τ] = d 2 d 2 τ τ θθ + τ φφ [ τ] θ = ηµ θ θ τ θθ ηµ θ τ φφ σφ θ = = 4µC 4 + 4µC 4 = 8.d.2 2µC σφ θ + 2µC σφ θ = 8.d.3 Since [ τ] =, viscous foces can be neglected in this flow and the Eule equation fo inviscid fluids may be diectly used. 28