Issuu on Google+

Boundary-layer growth on a rotating and accelerating sphere G. T. KARAHALIOS A N D V . THEOFILIS Depctrttt1er1t of' Physics, Uriii*er..sityof' Pntrcr.~,Ptrrrcis 261 10, Greece

Rcccivcd May 2 , 1986 Boundary-layer growth on a sphcrc is studied whcn it is sct into motion with constant acceleration and constant angular velocity, the latter being normal to the formcr. Analytic cxprcssions are derivcd for the velocity componcnts of the incompressible fluid in terms of a power series of the time of motion as well as for thc skin friction. On Ctudie la croissance dc la couchc liniite dc fluide sur unc sphbrc niisc en mouvemcnt avec une accClCration constante ct une vitessc angulaire constante, la secondc dtant perpendiculairc B la premiere. On dtablit des expressions analytiqucs pour les composantes de la vitesse du fluide inconipressible en termcs dc sCrie de puissances de la durCe du mouvcment, ainsi que pour le frottement pelliculairc. [Traduit par la revue] Can. J . Phys. 65. 23 (19x7)

1. Introduction Impulsive or accelerated motion of a body of revolution is of great importance in fluid mechanics, and it has been studied extensively. lllingworth ( I ) considered the boundary-layer growth on a body moving along its axis of symmetry and spinning about the same axis. Squire (2) treated the case of the impulsive translational motion of a three-dimensional body by employing the method of successive approximations to produce a time-series solution, which was also used by Goldstein and Rosenhead (3) in their description of the initial stages of motion of a cylinder. Wadhwa (4) determined the boundarylayer growth on a body of revolution that simultaneously has axial and angular acceleration. Karahalios ( 5 ) described the time-dependent formation of the boundary layer on a body of revolution having translation and spin, placing emphasis on the case of a sphere. In the present work, a sphere is moving with constant acceleration while at the same time it starts to rotate impulsively about an axis normal to the former direction. The object is to study the boundary layer, derive analytic expressions for the skin friction, and also determine the conditions for which skin friction becomes zero. ~

[2]

-au+ u - + au - - + wu- a+v- - - v - av at a[ r a e ag

uvdr r d[ +

-

at

a'u

ag2

ue v a v +

ra0

,

UVdr rd5

and the continuity equation is d(ru)

du

a(rw) d(

+-+-=o , i a0

Here U and V are the velocity components in the 5 and 0 directions, respectively, that the main stream would have in an incident stream with velocity Q in the negative x direction,

-

2. Statement of the problem and governing equations At t = 0, a sphere of radius R starts to rotate impulsively about, say, the z axis with a constant angular velocity R . Simultaneously, it starts to move along the x axis with velocity Q varying with time according to the equation Q = Dt, where D is the acceleration (see Fig. 1). The frame of reference O(x,y,z) moves forward with the sphere but does not rotate. We consider a local frame of reference ([,0,<), where [ is measured along a meridian curve from the pole, 0 is the azimuthal angle, and [ measures distance, along the outward normal, from the surface of the sphere. Hence, a point P is defined by the coordinates 5, 0, and [. The scale factors are hk = 1 , h,j = r , and hi = 1, where r is the radius of the cross section through P, normal to the z axis. Let 11, u, and w be the components of the fluid velocity. Then, the equations of motion of the boundary layer for incompressible flow are

& at

+

lr&

a[

+

u

rao

+

& , -

V

3 5 ,. -Q sin - sin 0 2 R

=

3. Approximate power-series solution Equations [ I ] and [2] can be simplified if we assume a solution with successive approximations of u , u, and w, each velocity component expressed in a power series of the time t. Initially, the effect of the impulsively started rotational motion is much more significant than that of the translational motion. , [2], we get Hence, if we put u = u ~ in

By setting uo = R R ( 1 - go) sin (SIR), where go is a function of the similarity variable 4

rl=-

2(v t)"' we obtain from [4], g;'

+2

~

o

=~

;

-

-

where primes denote differentiation with respect to -q, and the boundary conditions are go(0)

=

0,

go(m) = 1

The solution of [4] is

r d~;

aU +

at

5

cos -, R

vfi

- u'dr

ag

3 2

(/ = - -Q

u-a U + -V a U - -V' d r a[

r a0

r d(

where a = n-I". By setting u = u , in [I], we find that [ I ] takes the form


C A N . J . PHYS. VOL. 65.

1987

-2000 0

180

8

360

(deg) . )

FIG. 3. Variation of skin friction with the azimuthal angle 0 at the equatorial plane. N = 50.

FIG. I. Notation.

FIG. 2. Variation of skin friction with the azimuthal angle 0 at the equatorial plane. N = 10.

[5]

v-

aZuI

ag2

-

aul at

-= a ,

FIG. 4. Relation between distance .r:r and number N of revolutions before the skin friction becomes zero.

+ a z ( l - E)'

where

and

5 sin. -5 a 2 = -.R2R cos R R We seek a solution of [5] of the form ul

=

[(alfl + azfd

Hence, the following equations are derived:

[6]

f l + 2qf; - 4fl

[7]

fz

=

4

+ 2qf; - 4f2=4(1

with boundary conditions f , ( O )

- E)' =

0 , fl(co)

= 1,

f2(0) = 0 ,

FIG. 5. Variation of distance x* with angle 0 for zero skin friction.


25

KARAHALIOS A N D THEOFlLlS

and fi(m)

0. The solutions of [6] and 171 are

=

[8]

fl

=

2q'-

[9]

f2

=

2q2E2 E [ 4 ( a 2 - 1)q2 2'1' 4qae-"'1 2q2(1 - 2 a 2 ) - 2 a 2 4'1e-"'('1' - I)? + 2'I'e-'"1

+

+

We now put u

E(I + 2 q 2 ) - 2qae-"'

=

uo

+ +

+

b4

=

5 . 5 - R 3 ~ ( 2cos- - sln - - sin' R R

Putting

+ u l and u = u l in 121. Therefore, in [9], one finds

a vg ,,

or, if we use the expression u , g;'

+ 2qg;

-

4g,

=

= tx

with boundary conditions f 2 , ( 0 ) = f2,(m) = 0. In the same manner, by setting

4

,

with the boundary conditions g (0) = 0 and g , (30) = 1 and the solution given by [8]. We can calculate w , ,the first approximation to vv, from [3] as l t d ( u I r ) a2v d(a2r) vvl = -2(vt)I1'; [a t +-h, aeat at h

,I

+-

For the second approximation we put u

=

+ 141,

ul

u

2 bigi i= l

in [I I], we derive

with the boundary conditions gzi(0) = g2;(m)= 0, where s 2 ,= (1 - E ) f l , s2?=f 2 ( l - E), sz3= h,ae-"', and S74 = 2 h 2 a e - ~ ' . Equation [I31 has a solution of the form

where

+ u , + u2, and w = w, + w 2 in [ l ] and [2] to obtain

4

vz = t 2

=

uo

+

+

+

+

where tn = 4q4 12q2 3 and q = mE [l(4q3 10q)e-~' are the complementary functions, while the solution of [I21 is f ? l = g l l . Expressions for h , h 2 , X2,, etc. are presented in the Appendix. The skin friction components are given by

,

where b,

=

5 2 a , R cos R

bz

=

5 2a'R cos R

[IS] q [I61

=

T,, =

1 iP(T)

11'

1

11'

=

5 cos 0 sin R Substitution of the values of u , , u2, etc. into these expressions gives

3 cos X t [ I b f ; ( ~ ) - R2R sinX fi(0)

ip(;){-OR -

where X

+ -23R b

sin X g;(O)

3

+ t-b2

+ 3 R b sin X sin 0 tfiI(0)]

. sln X sin 0 g;(O)

~ R ' RCOS'X sin X gi2(0) + ( 3 R b cos 2X

3 + -fib 2

+ t 2 [ ( i R b cos2X + R23

b sin X cos 0)gi,(O)

sin X cos 0) g4,(0) - R'R(2 cos'X - sin2X) sin X g1,(0)

[/R and the values of gl(0), f l ( 0 ) , etc. are tabulated in the Appendix.

11

-

4. Results and discussion For X = 1 ~ 1 2 [15] , gives T~ = 0, a result also occurring in the case of impulsive translational [notion with spin (5), while [I61 gives, for the same value of X, I 11' [I71 T,, = p ( - ) OR(- 1.12838 - 1 . 0 7 7 5 2 ~sin 0 - 17.57335~' cos 0 - 1.12459x* + 1 .08447N2) N 2 t where N is the number of revolutions and x* [IS] x * =

(1.12459

+

=

4 bt2/R. It is, therefore, deduced that T,, becomes zero for

1.08447N2 - 1.12838 N 17.57335 cos O)N + 1.07752 sin 0


CAN. J . PHYS. VOL. 6 5 , 1987

26

Hence, for 0 = ~ 1 2and , for large values of N, the number x* of radii travelled by the sphere before T, becomes zero is large. The previous results are illustrated in Figs. 2-5. In Figs. 2 and 3, the skin-friction component T,, given by [17], has been plotted against the azimuthal angle 0. The comparison of the two diagrams shows that when the number of revolutions is increased, then the nondimensional distance x:': that has to be travelled for TO to become zero or negative increases. In Fig. 4 , the nondimensional distance x:"ersus the number N of revolutions ([I 81) has been sketched at a given azimuthal angle, 0 = 7~12.It can be deduced that the variation of x:"ith N is almost proportional. Evidently, when the number of revolutions is large while the time of motion is small, it is not possible for TO to become zero. Finally, in Fig. 5, x* versus 0 ([18]) has been plotted for various values of N. This drawing shows again that an increase in the value of N also drastically increases the value of s* for which T,, could be zero.

5. Conclusions When a sphere is simultaneously moving with constant acceleration and rotating with constant angular velocity, the rotational motion affects the length of time it takes for the skin friction to become equal to zero. In particular, an increase in the angular velocity increases the distance that has to be travelled by the sphere for the skin friction to reach a value of zero. If this is combined with the fact that the forward velocity approaches an upper limit, this suggests that practically speaking, one could avoid zero friction in accelerated translational motion by giving the sphere a large angular velocity normal to its direction of motion. I . C. R. ILLINGWORTH. Philos. Mag. 45, 1 (1954). 2. L. C. SQUIRE. Philos. Mag. 45, 1272 (1954).

3. S. GOLDSTEIN and L. ROSENHEAD. Proc. Cambridge Philos. Soc. 32, 392 (1936). 4. Y . D. WADHWA. Philos. Mag. 3, 152 (1958). 5. G. T. KARAHALIOS. Q. J . Mech. Appl. Math. 38, 389 (1985).

Appendix The coefficients of the error function in (141 are as follows:

whereA = 2 a ' -

1 andB

=

2(1

-

a').

g21:

+2

~

~

' + 2a'

- 1 1 ~ 2

1 -Ma2(4q' 2 8

- -

+

12q'

+ 3) erf f i q


KARAHALIOS A N D TkIEOFILIS

g23:

I Y2?= --(4~j' 24

Xz3 = 0 ,

40' Y~~ = -(7 45

tiz4=

4a2),

-

+

12~j' + 3 )

--

24

I + -a1(32$ 90

- 71)

The first derivatives of the functions f and g at 71 = 0 are

g;

2a

=

=

19 g 2, , = -a 3 giz gi3

=

=

64 + -a'

=

--a49 9

-

27 1 a' 180

--

gi,

-

9

35 gi4 = --a 18 f;

1.12838,

'7a8' 3

g; 8

=

-

=

-2.25676

2.16325

+ 6 \ / j a 3 = -3.69453

-

9

+ 23 = 0.18743 -

256 + -a'855 + -a5 45 45 fi

-40

=

=

-8(1 - a')

-

16cr'-2 1 f i 40

=

+ 4 a = -0.82006

0.02747


1987__KarahaliosTheofilis__CanJPhys_Vol65_pp_23-27_1987