KUNDU Fluid Mechanics 2 Episode 8 pot

It is clear that low Reynolds number flows will have ncgligible inertia forces and therefore the viscous and pressure forces should be in approximate balancc.. 9.60 were multiplied by Re

3 Difliion of a V i r h Sheet

Consider the case in which the initial velocity field is in the form of a vortex shcct with u = U €or y > 0 and u = -U for y < 0 We want to investigate how the vortex sheet decays by viscous dflusion The governing equation is

where sgn(y) is the "sign function," defined a, 1 €01 positive y and -1 for negative

Y As in thc previous section, the parameter U can be eliminated €om the governing set by regarding u / U as the dependent variable Then u / U must bc a function of

(y, t, v), and a dimcnsional analysis reveals that there must exist a similarity solution

(9.38)

A plot of the velocity distribution is shown in Figure 9.11 If we define the width of

h e transition layer as the distance between the points where u = f0.9SU, then the corresponding value of r,~ is f 1.38 and consequently the width of the transition layer

is 5.52,';i

It is clear that the flow is essentially identical to that duc to the impulsive start

of a flat plate discussed in the preceding section In fact, each half of Figure 9.1 1

is idcntical to Figure 9.10 (within an additive constant of f l ) In both problems

290 hfflinar I;yuIL

Figure 9.11 Viscous decay of a vortex shect Thc right panel shows thc nondimcnsional solution and thc

left panel indicatcs h c vorticity distribution at two tirncs

the initial delta-function-like vorticity is diffused away Tn the presenl problem the magnitude of vorticity at any time is

(9.39) This is a Gaussian distribution, whose width increases with timc as a, while the maximum value decreascs as I/& The total amount of vorticity is

which is independcnt of time, and equals the y-integral of the initial (delta-function-like) vorticity

9 Decay of a Line h r k x

In Section 6 it was shown that when a solid cylinder of radius R is rotated at angu-

lar specd s2 in a viscous fluid, the resulhg motion is irrotational with a velocity distribution U S = ! 2 R 2 / r The velocity distribution can be writkn as

r

U o = - 9

wherc r = 2n SZ R2 is thc circulation along any path surrounding the cylinder Suppose

the radius of the cylinder goes to zero while its angular velocity correspondingly

2x1-

inmeases in such a way that the product r = 2irQR' is unchanged In the limit we

obtain a line vortex of circulation r, which has an infinite velocity discontinuity at thc origin

Now suppose that the limiting (infinitely thin and fast) cylinder suddenly stops rotating at r = 0, thereby reducing the velocity at the origin to zero impulsively Then the fluid would gradually slow down from the initial distribution because of viscous diffusion from the region near the origin The flow can therefore be regarded as that of the viscous decay of a line vortex, for which all the vorticity is initially concentrated

at the origin The problem is the circular analog o€ the decay of a plane vortex sheet discussed in the preceding section

Employing cylindrical coordinates, the governing equation is

subject to

(9.42) (9.43)

We expect similarity solutions here because there are no natural scales for Y and t introduced from the boundary conditions Conditions (9.41) and (9.43) show that the dependence of the solution on the parameter r/21rr can be eliminated by defining a nondimensional velocity

(9.44)

which must have a dependence of the form

u' = f ( r , t , u )

As thc lcft-hand side of the preceding equation is nondimensional, the right-hand side

must be a nondimensional function of r, t, and u A dimensional analysis quickly

shows that the only nondimensional group formed from thcsc is r/Jvb Therefore, the problem must have a similarity solution d the form

u' = F ( q ) ,

(9.45)

(Notc that we could have defined q = r/2& a$ in the previous problems, but the

algebra is slightly simpler if we define it as inEq (9.45).) Substitution of thc similarity solution (9.45) into the governing set (9.40X9.43) givcs

F" + F' = 0,

subject to

F ( 0 ) = 1,

P ( 0 ) = 0

292 Imminar Flow

r Rprc! 9.12 Viscous dccrry of a line vortcx showing the iangcnlial velocity at diJTcrent times

The foregoing discussion applies to the &cay of a line vortex Consider now the case where a line vortcx is suddenly introduced into a fluid at rest This can be visualized as the impulsive start of an infinitely thin and fast cylindcr It is easy to

show that the velocity distribution is (Exercise 5 )

(9.47)

which should be compared to Eq (9.46) The analogous problem in heat conduction

is the sudden introduction of an infinitely thin and hot cylinder (containing a finite

amount of heat) into a liquid having a different tcmperature

10 Flow llue to an Oscillahg Plate

The unsteady parallel flows discussed in the three preceding sections had similarity solutions, because there were no natural scales in space and time We now discuss

IO J h u Ilue to an 0.wilhtitig !'hie

an unsteady parallel flow that does not have a similarity solution bccause of the

existence ora natural time scale Consider an idmite flat plate that executes sinusoidal

oscillations parallel to itself (This is sometimes called Stokes' secondproblem.) Only

the steady periodic solution a~let- the slarting transients have died will be considcred,

thus there are no initial conditions to satisfy The governing equation is

293

subject to

u(0, t) = u cos w t , u(00: r) = bounded

(9.48)

(9.49) (9.50)

In the stcady statc, thc flow variables must have a periodicity equal to the periodicity

of the boundary motion Consequently, we use a separable solution of the form

where what is meant is the real part of the right-hand side (Such a complex form

of represcntation is discussed in Chapter 7, Section 15.) Here, f ( y ) is complex,

thus u ( y , t) is allowed to have a phase difference with the wall velocity U cos w l

Substitution of Eq (9.51) into the governing equation (9.48) gives

(9.52)

This is an equation with constant coefficients and must have exponential solu-

tions Substilution of a solution of the form f = exp(ky) gives k = m =

&(i + l)-, where the two square roots of i have been used Consequently,

the solution of Eq (9.52) is

(9.53)

The condition (9.50), which requires that the solutionmustremain boundcd a1 y = 30,

needs B = 0 The solution (9.51) then becomes

= A e i w ~ , - ( l + i ) y , h P (9.54)

The surface boundary condition (9.49) now givcs A = U Taking the real part of Eq

(9.54), we finally obtain the velocity distribution for the problem:

The cosine term in Eq (9.55) represents a signal propagating in the direction of

y , while the exponcntial term represents a dccay in y The flow thercfore resem-

bles a damped wave (Figure 9.13) However, this is a dfision problcm and nor a

294 imminar I-liiw

U

Figore 9.13 Velocity dishbution in laminar flow near an osdllating plalc The distributions at wf = 0,

x / 2 , n, and 3n/2 are shown Thc dillilsive distmcc is of order d = 4 m

wave-propagation problem because there are no rcstoring forces involved here The apparent propagation is merely a result of the oscillating boundary condition For

y = 4 m , ihc amplitude of u is U exp(-4/&) = O.O6U, which means that the influence of the wall is confined within a distance of order

which decreases with frequency

Note that the solution (9.55) cannot be mpresented by a single curve in krms of

the nondimensional variables This is expected because the frequency of the bound- ary motion introduces a natural time scale l/o into the problem, thereby violating the requiremcnts of self-similarity There are two parameters in the governing set

(9.48)-(9.50), namely, U and w The parameter U can be eliminated by regarding

u / U as the dependent variable Thus the solution must have a form

U

- = f ( Y , t , 0: V I (9.57)

As there are fivc variables and two dimensions involved, it follows that there must be

three dimensionless variables A dimensional analysis of Eq (9.57) gives u / U , of,

and y mas the three nondimensional variables as in Eq (9.55) Self-similar solu- tions exist only when there is an absence of such naturally occurring scalcs requiring

a reduction in the dimcnsionality of the space

An interesting point is that the oscillating plate has a constant diffusion dis-

tance 6 = 4 m that is in contrast to the casc of the impulsively started platc

U

in which the diffusion distance increases with time This can be understood from the govcming cquation (9.48) In thc problcm of sudden accelcration of a plate,

i12u/i)y2 is positive for all y (see Figure 9.10), which results in a positive au/at

everywhere The monotonic acceleration signifies that momentum is constantly diffused outward, which results in an ever-increasing width of flow In contrast,

in thc casc of an oscillating plate, a2u/i3y2 (and therefore a u / a r ) constantly changes sign in y and t Therefore, momentum cannot diffuse outward monotonically, which results in a constant width of flow

The analogous problem in heat conduction is that of a semi-infinite solid, the surhce of which is subjected to a periodic fluctuation of temperature The resulting solution, analogous to Eq (9.59, has been used to estimate the effective “eddy”

diffusiviry in thc upper layer of the ocean from measurements of the phase difference

(that is, h e time lag between maxima) between the temperature fluctuations at two depths, generated by the diurnal cyclc of solar heating

11 Hifih and 1,ow Reynolds :I:Wnber 1~’Lowx

Many physical problems can be describcd by ihe behavior of a system when a certain parameter is either very small or very large Consider the problem of steady flow around an object dcscribed by

First, assume that the viscosity is small Then the dominant balance in thc flow is between the pressure and inertia forces, showing that pressure changcs are of order

p U 2 Consequently, we nondimensionalize the governing cquation (9.58) by scaling

u by the frcc-strcam velocity U , pressure by p U 2 , and distance by a representative

lcngth L of the body Substituting the nondimensiond variables (denoted by primcs)

the equation of motion (9.58) becomes

where Re = U L v is thc Reynolds number For high Reynolds number flows,

Eq (9.60) is solved by treating 1/Re as a small parameter As a h s t approxima- lion, we may set 1/Re to zero everywhere in thc flow, thus reducing Eq (9.60) lo

the inviscid Euler equation However, this omission of viscous terms cannot be valid near the body because thc inviscid flow cannot satisfy the no-slip condition at the body surface Viscous forces do become important near the body becausc of the high

shcar in a layer near the body surfacc The scaling (9.59), which assumes that veloc- ity gradients are proportional to U/L, is invalid in thc boundary layer near the solid surface We say that there is a region of nonunifornib): near the body at which point

a perturbation expansion in terms of the small parameter 1 /Re becomes singulur

The proper scaling in the boundury luyer and the procedure of solving high Reynolds number Rows will be discussed in Chapter 10

296 Laminar Flow

Now consider flows in the opposite limit of very low Rcynolds numbers, that

is, Re + 0 It is clear that low Reynolds number flows will have ncgligible inertia forces and therefore the viscous and pressure forces should be in approximate balancc For the governing equations to display this fact, we should have a small parameter

multiplying the inertia forces in this case This can be accomplished if thc variables

are nondimensionalized properly to take into account the low Reynolds number nature

of the flow Obviously, the scaling (9.59), which leads to Eq (9.60), is inappropriatc

in this case For if Q (9.60) were multiplied by Re, then the small parameter Re would appear in front of not only the incrtia force term but also the pressure € m c

term, and the governing equation would reduce to 0 = pVzu as Re + 0, which is

not thc balance for low Reynolds number flows Thc source of the inadequacy of the nondimemionalization (9.59) for low Reynolds number flows is that thc pressure is

not of order p U 2 in this case As we noted in Chapter 8, for these extcrnal flows, pressure is a passive variable and it must be normalized by the dominant efFcct(s), which here are viscous forces The purpose of scaling is to obtain nondimensional variables that are of order one, so that pressure should be scaled by p U z only in high Reynolds number flows in which the pressure forccs are of the order of the inertia

forces In contrast, in a low Reynolds numbcr flow the pressure forces are of the order

of the viscous forces For V p to balance p V z u in Eq (9.58), the pressure changes

must have a magnitudc of the ordcr

p - L p P u - p U / L

Thus the proper nondimensionalization for low Reynolds number flows is

(9.61)

The variations of the nondimensional variables u‘ and p’ in the flow ficld are now

of ordcr one The pressure scaling also shows that p is proportional to p in a low Reynolds number flow A highly viscous oil is used in the bearing of a rotating shaft because the high pressure developed in the oil film of thc bearing “lifts” the shaft and prevents metal-to-metal contact

Substitution of Eq (9.61) into (9.58) gives the nondimensional equation

Re uf Vu’ = -Vp’ + v2u’ (9.62)

In the limit Re + 0, Eq (9.62) becomes the linear equation

where the variables have been converted back to thcir dimensional hm

Flows at Re << 1 are called creeping motions They can bc due to small velocity,

large viscosity, or (most coinmonly) the small sizc of the body Examplcs of such flows are the motion of a thin film of oil in the bearing of a shaft, settling of sediment particles near the ocean bottom, and the fall of moisture drops in the atmosphere In thc next section, we shall examine the creeping flow around a sphere

Sumrmri-y: The purpose of scaling is to generate nondimensional variables that are of order onc in the flow field (except in singular regions or boundary layers)

The proper scales depend on the nature of theJlav a d are obtained by equating the terms thut are most important in the flow field For a high Reynolds number

flow, thz dominant terms are the inertia and pressure forces This suggests the scaling

(9.59) resulting in the nondimensional equation (9.60) in which the small parameter multiplies the subdominant term (except in boundary layers) In contrast, the dominant terms for a low Reynolds number flow are the pressure and viscous forces This suggests the scaling (9.611, resulting in the nondimensional equation (9.62) in which the small parameter multiplies the subdominant term

12 &?c?ping Flow murid a Sphere

A solution for the creeping flow around a sphere wa, iirst given by Stokes in 185 1

Consider the low Reynolds number flow around a sphere of radius a placed in a uni- form stream CJ (Figure 9.14) Thc problem is axisymmetric, that is, the flow patterns

are idcntical in all planes parallel to U and passing through the center of the sphere Since Re + 0, as a first approximation we may ncglect the inertia forces altogether and solve the equation

We can form a vorticity equation by taking the curl of the preceding equation, obtain- ing

Here, we have used the fact hat thc curl of a gradient is zero, and that the order of thc operators curl and V2 can be interchanged (The reader may verify this using indicia1 notation.) The only component of vorticity in this axisymmetric problem is q,, the component perpendicular to (p = const planes in Figure 9.14, and is given by

The governing equation is

Combining the last two equations, we obtain

v2w, = 0

2

[$+-+-)I sin0 a 1 a $ = O

r2 ae sin0 a0

298 Larninurlylrw

-re 9.14 Creeping flow ovee a sphcrc The uppmpanel shows lhc blur slrars componena at the

sllrke n l c l o w c r ~ shows h ~ d i s t r i h t i r n in rmanial (p = amst.) plane

The boundary conditions on the preceding equation ~IE

Ma, e) = 0 [u, = o atsurface], (9.65)

*(oo, e) = ;ur2 sin2 e [uniform at 001 (9.67)

The last condition follows from the fact that the stream function for a uniform flow

is (1/2)Ur2 sin2 8 in spherical coordinates (see Eiq (6.74))

ag/ar(a, e) = 0 [ue=O atsllrfacel, (9.66)

The upsheam condition (9.67) suggests a separable solution of the form

The upstream boundary condition (9.67) r e q u k s that A = 0 and B = U/2 The

surface boundary condition then gives C = -3 Ua/4 and D = Ua3/4 The solution

(9.68)

The velocity components can thcn bc found as

Thc prcssure distribution is sketched in Figurc 9.14 The pressure is maximum at

thc forward stagnation point where it equals 3 p U / 2 a , and it is minimum at the rcar stagnation point where it equals -3pU/2a

Let us determine the drag force D on the sphere Onc way to do this is to apply the principlc or mechanical energy balancc over the entire flow ficld givcn in Eq (4.63)

This requires

D U = # d V ,

which statcs that the work done by the cylindcr equals the viscous dissipation over the entirc flow; hcre, # is the viscous dissipation per unit volume A morc direct way

to dctciminc the drag is to integrate thc stress over the surfacc of the sphere The force

per unit area normal to a surhce, whose outward unit normal is n is

s

F; = t ; , n j = [ - p G i j + q ] n , = -pni + o n 11 I ?

where t i j is thc total stress tensor, and o;,j is the viscous strcss tcnsor The component

of the drag rorcc per unit area in thc direction of the uniform stream is thereforc

[ - p cos 8 + or, cos H - ore sin O],.,, , (9.71)

which can be understood from Figure 9.14 TIic viscous stress componcnls are

300 r h R h

of which onethird is ptssure drag and two-thirds is skin fiction drag It follows that the resistance in a creeping flow is proportional to the velocity; thk is known as

Stokes’ law ofresisrance

In a well-known experiment to measure the charge of an electron, M i l l h used

Eq (9.73) to estimate the radius of an oil drapret falling through air Suppose p’ is the density of a spherical falling particle and p is the density of the surrounding fluid

Then the effective weight of the sphere is 4nu3g(p‘ - p)/3, which is the weight of

the sphere minus the weight of the displaced fluid The falling body is said to reach

the ‘‘termid velocity” when it no longer accelerates, at which point the viscous drag

equals the effective weight Then

+l3g(p’ - p) = 6sfjMu,

h m which the radius a can be estimated

Millikan was able to deduce the charge on an electron making use of Stokes’ drag

f o d a by the following experiment ’ k o horizontal parallel plates can be charged

by a battery (see Fii 9.15) Oil is sprayed through a very fine hole in the upper plate and develops static charge (+) by losing a few (n) electrons in passing through the

mall hole IF the plates are charged, then an electric force neE will act on each of

the dmps Now n is not known but E = -V’/L, where Vj, is the battery voltage

and L is the gap between the plates, prmrided that the charge density in the gap is very low With the plates uncharged, measmment of the downward terminal velocity

allowed the radius of a drop to be calculated assuming that the viscosity of the drop

is much larger than the viscosity of the air The switch is t h w n to charge the upper plate negatively The same droplet then reverses direction and is f o r c e d upwards It quickly achieves its terminal velocity Vu by virtue of the balance of upward forces

(electric + buoyancy) and downward forces (weight + drag) This gives

6sfpUua + (4/3)nu3g(p‘ - p ) = neE,

where U,, is measured by the obseMltion telescope and the radius of the particle is

now known The data then allow for the calculation of ne As n must be an integer,

data from many droplets may be Merenced to identify the minimum difFerence that

must be e, the charge of a single electron

The drag coefficient, defined as the drag force nondimensionalized by pU2/2

and the projected are xu’, is

(9.74)

where Re = h U / u is thc Reynolds number based on the diametcr of the spherc

In Chapter 8, Section 5 it was shown lhat dimcnsional consideralions alone rcquire that C D should be inversely proportional to Rc for creeping motions To rcpcat the argument, the drag force in a “massless” fluid (that is, Re << 1) can only have the dcpendence

D = f ( p , U , a)

The preceding relation involves four variables and thc t h m basic dimensions of mass,

length, and timc Therefore, only one nondimcnsional parameter, namely, D / p U u ,

can be formed As thcrc is no second nondimensional parameter for it to depend on,

D / p U a must be a constant This lcads to C D a I /Re

Thc flow pattern in a reference frame fixed to the fluid at infinity can be found

by superposing a uniform velocity U to the left This cancels out the first term in

Eq (9.68), giving

$ = U r s i n Q * [ +- : f 3 ] , which givcs the streamlinc pattern as sccn by an obscrver if the sphcre is dragged

in front of him from right Lo left (Figurc 9.16) The paltern is symmctric between

F i y e 9.16 Strcamlincs and vcltrity distributions in Stokcs’ solution ofcwcping flow duc u) a moving

sphcre Yore thc upslrcam and downstream symmcwy, which is a result ofcornplek neglca ornonlinenrity

the upstream and the downstream directions, whicb is a result of the linearity of the

governing equation (9.63); reversing the dimtion of the f e s t r e a m velocity merely changes n to -n and p to p The flow therefore does not have a “wakc” behind the

This shows that the inertia farces are not negligible for distaaccs huger than r / a -

l/Re At sufsciently large distances, no matter how small Re may be, the neglected

Solutions of problems involving a small parameter can be developed in terms

of the perturbation series in which the highcr-order terms act as conrecrions on the

lower-order terms perturbation expansions are discussed briefly in the following chaptcr If we r c g d the Stokes solution as the first term of a series expansion in the

small parameter Re, then the expansion is ‘n0nuniforrn’’ because it b d down at

a n i t y If we tried to calculate the next term (lo order Re) of the perturbation series,

we would find that the velocity corresponding to thc h i g h e r d r term bccomes unbounded at infinity

The situation becomes worse for two-dimensional objects such as the circular

cylinder Tn this case, the Stokes balancc V p = pV2u has no soluiion at all h a t can

satisfy the uniform flow boundary condition at infinity From this, Stokes concluded that steady, slow flows around cylinders canmt exist in M~UIC Tt ha now becn mdizcd

that the nonexistence of a h t approximation of the Stokes flow around a cylinder is due to the singuzlrr nature of low Reynolds number flows in which there is a region

of nanunifurmity at infinity The nonexistence of the second approximation far flow around a sphere is due Lo the same reason In a different (and more f a m i b ) class

of singular jxrturbation problems, the rcgion of nonuniformity is a thin layer (the

“boundary layer‘? near ihe surface of an object This is the chss of flows with Re +

00, that will be discussed in the next chapter For the= high Reynolds numbcr flows the small parameter 1/Re multiplies the higkst-order derivative in the governing

equations, so that the solution wilh ]/Re identically set to zero cannot satisfy all terns became arbitrarily large

the boundary conditions Tn low Reynolds number flows this classic symptom of

the loss of the highcst derivative is absent, but it is a singular perlurbation problem nevertheless

In 1910 Oseen provided an improvement to Stokes‘ solution by partly accounting for the inertia terns at large distances He made the substitutions

I

u = u + u ’ v = u ‘ w = w :

where (u’: u’, w’) arc thc Cartcsian componcnts of the pcrturbation velocity, and arc small at large distances Substituting these, the advection term of the x-momentum cquation becomes

Neglecting the quadratic terms, the equation of motion bccomcs

where ui represents u’, v’, or w’ This is called Oseen’s e y u d o n , and the approxima-

tion involved is called Oseen’s approximation Tn essence, the Oseen approximation linearizes h e advective term u h by U(au/ax), whcreas the Stokes approximation drops advection altogether Near the body both approximations havc the same order

of accuracy However, the Oscen approximation is better in the far field where the velocity is only slightly different than U Thc Oseen equations provide a lowest-order solution that is uniformly valid cverywhere in the flow field

The boundary conditions for a moving sphere arc

u’ = li’ = UI’ = 0 at infinity

u’ = -U, V I = w’ = 0 at surface

The solution found by Oseen is

(9.75)

3

l/a2 $ = [;I2 - + - l r ] sin'^ - Re -(I +cos e)

wherc Re = 2 a U / v is the Reynolds number based on diameter Ncar the s d a c e

r / a rz 1, and a series expansion of the cxponential term shows that Oseen’s solution

is identical to the Stokes solution (9.68) to the lowest order Thc Oseen approximation predicts that the drag coefficient is

The Streamlines -

to the oseen solution (9.75) are shown m

Fgr 9.17, where a d o r m flow of U is added to the left so as to generate the

pat- of flow due to a sphere moving in front of a stationary obsemx It is seen

that the flow is no longer symmelrick but has a wake where the streamlines are closer

togeher than in the Stokes flow The velocities in the & arelargerthaninfront of

the sphere Relative tothe sphere, the flow is slower in the wake than i n h t of the

In 1957, Oseen’s Cmrection to Stokes’ solution was rationalized independently

by Kaplun and proudman and Pearson in terms of matched asymptotic expansions Here, we will obtain only the h t a d e r correction The full vorticity equation is

better approximation in the inner region, we will write

W, ,w Re) = $dr, PI + Re $1 (r, 1-4 + o ( R e ) , (9.78) where Khc sccond correction “ ~ ( R c ) ” means that it tends to zero faster than Re in the limit Re + 0 (See Chapter 10, Section 12 Here p? is made dimensionless by Uu2

and Rc = V a / u ) Substiluting Eq (9.78) into (9.77) and taking the limit Re -+ 0,

we obtain D4$o = 0 and recover Stokes’ result

Subtracting this, dividing by Re and taking thc limit Re -+ 0, we obtain

which reduces to

(9.79)

(9.80)

by using Eq (9.79) This has the solution

where CI is a constant of integration for thc solution to h e homogeneous equation and is to bc dctermined by matching with the outer region solution

Tn the outcr mgion rRe = p is fie The lowest-ordcr outcr solution must be

uniform flow Then wc write the streamfuntion as

I P 2 2 1

* ( p l 0; Re) = z- sin 0 + QI(p, 0) + o

Substituting in Eq (9.77) and taking the limit Re + 0 yields

where the opcrator

The solution to Eq (9.82) is round to be

(9.82)

where the constant of integration C2 is determined by matching in the overlap region between the inner and outer regions: I << r << 1/Re, Re << p << 1

The matching gives C2 = 314 and CI = -3116 Using this in Eq (9.81) for the

inner region solution, the O ( R e ) correction to the stream function (Eq (9.81)) has been obtained, fiom which the velocity components, shear stress, and pressure may

be derived Intcgcating over the surface of Lhe sphere of radius = a, we obtain the final result for the drag force

D = 6npUa[l + ~ U U / ( ~ V ) ] ,

which is consistent with Oseen’s result Higher-order corrcctions were obtained by

Chester and Breach (I 969)

14 Hde-Shaw Plow

Another low Reynolds number flow has seen wide application in flow visualization apparatus because of its peculiar and surprising property of reproducing the stream- lines of potential flows (that is, infinite Reynolds number flows)

The Hele-Shaw flow is flow about a thin object filling a narrow gap between

two parallel plates Let the plates be located at x = f h with Re = U,h/v << 1 Here, UO is the velocity upstream in the central plane (see Figure 9.1 8) Now place a circular cylinder of radius = a and width = 2b between the plates We will require

b/a = E << 1 The Helc-Shaw limit is Rc << E‘ << I Imagine flow about a thin coin

with parallel plates bounding the ends of the coin We are interested in the streamlines

of the flow around the cylinder The origin of coordinates ( R , 8 , x ) (Appendix B) is

taken at the center of the cylinder

Consider steady flow with constant density and viscosity in the absence of body

forces The dimensionless variables are, x’ = x / h , R’ = r / a , d = v/U,, p’ = (p - p , ) / ( p U , / b ) , Re = U,b/v, E = b/a Conservation of mass and momentum

then take the following form (primcs suppressed):

With u , = O ( r ) at most, a p / a x = O ( c ) at most so p = p ( R , 0) Inkgrating the

momcnturn cquations with respect to x,

where no slip has bccn satisfied on x = f I Thus we can write u = V 4 For h e

~wo-diinensionalfielduK,u,,.Here,4 = - i p ( l -x2) Now werequirethatu, = O ( E )

so that the first term in thc continuity equation is small compared with the others Then

Substituting in terms of thc vclocity potential 4, we havc V24 = 0 in R, N subjccl lo

(no mass flow normal to a solid boundary)

The solution is just the potcntid flow over a circular cylinder (Eq (6.35))

1 (:I - X * )

4 = R C O S ~ ( 1 + - R 2 ) - 2 '

308 I m a k r f b

wherex is just aparameter Therefore, the sfrcamkx correspondingtothisvelocity

potential ace identical to the potential BOW s h r e m of Eq (6.35) This allows for

the ccmstmction of an apparam to visualize such potential flows by dyc injection

between two closely spaced glass PI- The velocity diyiributim of h i s flow is

As R + 1 , u ~ + Obutthcreisaslipvelocityue + -2YinO(l -x2)/2

A s t hi si s a~~u s flo w, the re mu s te xi s tathi nr eg ion ne ar R = lwhere

Lhc slip velocity ug decreases rapidly to ZLXO to salisfy U g = 0 on R = 1 This

Lhin boundary layer is ~ e r y close to the body snrface R = 1 Thus, U R a 0 and ilp/aR X 0 throughout the layer NOW p = - R c o s e ( l + 1/R2) SO for R W 1,

large so the dominant balance i s

( i / R ) a p / a e w 2sine ~n the e momentum quation, R e ~ t i v e ~ become ~ e r y

It is clear liom this balance that a stretching by 1 / ~ is appmPriate in the boundary

layer: i = ( R - I)/€ IU these terms

We conclude that HebShaw flow indeed shdates pomtial flow (inviscid) strcam-

lincx except for a vcry thinboundary layer ofthe order of the platc separalion adjacent

to the body surface

f- 309

a computer- A proper application of such techuiques q u i m considerable care and

familiarity w i h various iterative techniques and their limitations It ihoped that the

rcadcr will have the opportunity to learn numerical methods in a separate study In

Chapter 11, we willintroduce severalbask methods of computational fluid dynamics

k%

1 Consider the laminar flow of a fluid layer falling down a plane inclined at an

angle 0 with the horizontal lf h is the thickness of Lhe layer in the fully developed

stage, show that the velocity distributiun is

where the x-axis points along thc frcc surEacc, and Lhc y-axis points txrward h e plane

Show that the volume flow rate per unit width b

gh3 sin 8

Q = 3v I

and the friczional s m s on the wall is

to = pgh sine

2 Consider the steady laminar flow tfirough the annular space formed by two

coaxial tubes Thc flow is along the axis of the tubes and is maintaincd by a pressure

gradient dp/dx, where the x direction i s Mien along the axis of the tubes Shaw hat

the velocity at any radiu r k

where a is the radius of the inner tube and h is the radius ofthe oulcr Lube find the

radius at which the maximum velocity i s rcached, the volume rate of flow, and thc

strcss disbibution

3 A long vertical cylinder of radius b rotates with angular velocity R concen-

trically outside a smaller stationary cylinder of radius a The annular spam is filled

with fluid of viscosity p- Show that the steady velocity distributian is

r2 - a2 n2n

b2-a’ r

ug =

Show that the torgue exerted on cither cylinder, pcr unit lengtn, equals

4 con side^ a solid cyhdex of radius R, steadily rotaling at angular speed R in

Assumc Re = p U a / p << 1 Neglm surface tension

9 Consider a vcry low Reynolds number flow over a circular cyclinder of radius

r = a For r / a = O( 1) in the Rc = Ua/u + 0 Limit, find the equation governing the

streamfunction @(r, 0) and solve for $ with the least singular behavior for large r

There will be one rcmaining constant of integration to be determined by asymptotic

matching with thc large r solution (which is not part of this problem) Find the domian

of validity of your solution

IO Consider a sphere of radius r = u rotating with angular velocity w about a diametcr so that Re = w 2 / u << 1 Use the symmetries in the problem to solve the