Vol. 5, No. 1 HU et al.: A method for exploring some new solutions of complex lamellar flow 17

A method for exploring some new solutions of complex lamellar flow

Xin HU*, Yongnian HUANG’ and Yi ZENG** ‘Department of Mechanics and Engineering Science, State Key Laboratory for Turbulence Research, Peking University, Beijing 100871, China **Basic Department, Jiaotong University of Eastern China, Nanchang 330013, China

(Received March 10, 2000)

Abstract: Complex lamellar flow is a typical flow in fluid mechanics. Its research has high theoretical value. Generally speaking, there are only two typical flows in nature: one is Beltrami flow, the other is complex lamellar flow. We have known much about the former. But the solutions of the latter are hard to obtain because it is difficult to solve the N-S equation or Euler equation. From the incompressible condition, if we let the form of the velocity satisfy some special conditions, we would obtain some new solutions of this flow. Keywords: Beltrami flow, complex lamellar flow, exact solution, incompressible condition

Introduction

The key questions of fluid mechanics are focused on the researches of velocity u and vor- ticity w. Often we take the views of field to study their relationship. And the orthogonal decomposition of vector w with respect to the direction of u can be written as follows:

uxw w=++ ,u,2 xu=Xu-kHxu

When H = 0, the flow is Beltrami flow, which is defined by w = Xu; while X = 0, the flow is complex lamellar flow, which is defined by w . u = 0. So all the 2-D flows satisfying the N-S equation are the complex lamellar flows. The research of complex lamellar flow will do great help to our knowledge of understanding the complexity and generality of the flow. But such solution is hard to get because it is hard to solve the N-S equation or Euler equation.

From Helmholtz-Hodge decompositionl’l, u = VP + XVq. For complex lamellar flow, w . u = 0, we will get VP = aVX + bVq and Va x VX + Vb x Vn G 0. Usually, we take account of a = b = O[ls21. So u = XVq, and w = VX x Vr]. In the following sections, we will discuss the flow satisfying the steady Euler equation: (u . V)w = (w . V)u and the incompressible condition: V .u = 0 in rectangular coordinate system (5, y, z) and axisymmetric cylindrical coordinate system (T, 8, z) (a/&9 = 0).

1 Rectangular coordinate system

We let X and q satisfy the conditions as follows:

rl = f(x) + S(Y) + 4)

Here abc # 0. So velocity u and vorticity w are like the following:

u = abc(f’e, + g’ear + h’e,)

w = (acb’h’ - abc’g’)e, + (abc’f’ - bca’h’)e, + (bca’g’ - acb’f’)e,

18 Communications in Nonlinear Science & Numerical Simulation March 2000

where e,, ep, e, are the unit vectors of x, y, z axes, respectively. The incompressible condition is simplified to:

(f” + ;f’, + (g” + $1) + (h” + ;h’) = 0

Thus, we have the following equations: I

f” + ;f’ = Cl, b’

g” + p’ = c2, h” + ;h’ = c3 = -(cl + c2)

And the Euler equation can be reduced to:

l;f’(cb’h’ - bc’g’) + ch’[b”g’ - (cl - cs)b’] + bg’[(q - c2)c’ - c”h’] = 0

2;g’(uc’f’ - ca’h’) + .f’[c”h’ - (~2 - CI)C’] + ch’[(cz - c3)u’ - d’f’] = 0

2$‘(ba’g’ - ab’f’) + bg’[u”,’ - (c3 - c2)u’] + af’[(c3 - q)b’ - b”g’] = 0

Because of the symmetry of the rectangular coordinate system, there are only four situations for us to discuss:

1) If w1 # 0, ws # 0, and ws # 0, or cb’h’ - bc’g’ # 0, uc’f’ - cu’h’ # 0, and ba’g’ - ub’f’ # 0, we will finally (after some tedious operations) get:

f’=nl, g’=nz, h’=ns, a=Aeklz, b=Bek2y, c=CektZ

Here, w, 712, n3, ki, ks, ks are all constants, cl =niki, c2 = nzkz, q = n3k3, and nlkl + n2ka + nsk3 = 0. So

u = (nl& 1224 n3N, w = ((n&z -n&3)X, (nlh - n&l)& (n&l - nlk$)

Here, X = ABCekl++k2Y+k3Z.

2) If w1 = 0, while w2 # 0, ws # 0, we would have two more situations to discuss:

2.1) If b’ = c’ = 0, then we will get:

u = (-PX2, PXY, PXZ), w = (0, -I=, PY)

2.2) If b’ # 0, c’ # 0, we will then have:

g’ = n2, h’ = ns, b = Bek2Y, c = CeksZ

Here, cs = n2k2, cs = risks. We will also have a’ = muf’, and m = ks/ns = ks/ns, so we have cl = -m(ng + ni). Let us introduce two constants ni and kl, which satisfy ni + nz = ni and m = -kl /ni , so cl = ni ICI. After some operations, we will have:

f” + m(f’)2 = cl = nlkl, a” = clmu = -m2n:u = -kfa

Soa=Acoski(x-xs),thenf’=nitanki(x-xo),so

u = (nl sin ICI (x - xo)H, n2 cos kl (x - zo)H, n3 cos kl (x - xo)H)

Here, H = ABCek21+kSZ, and nlkl + n2k2 + nsk3 = 0.

3) If wi = w2 = 0, while ws # 0, then we would have:

c = const, h = const, c3 = 0, cl + c2 = 0

Vol. 5, No. 1 HU et al.: A method for exploring some new solutions of complex lamellar flow 19

In fact, it turn to be the 2-D situation (z, y). And we will have the equation below:

a” a’ b” b’ a-~17= -++I--=m=const

af b W (2)

We must bring (2) and (1) together to solve the velocity. We could obtain some special solutions when u” = &/$a, and b” = fkzb. It is easy to get velocity solution at this time. Here we bypass the results because of the limit of the article’s length.

4) If w1 = w2 = w3 = 0, we would obtain:

a’ b’ c’ -z-x- af’ bg’ ch’

= A = const

a” I, -=B+(~c~+cI)A,

b - = B + 3~2.4,

a b c” = B + (c2 - q)A

C (3)

f” + Aff2 = cl, g” + Agt2 = c2, h” + Ah’2 = -(cl + ~2) (4)

It only needs to analyze the sign of the parameters A, cl, cz, etc. to obtain the solutions. For example, for f” + Af’2 = cl, you can divide it into six different situations to discuss:

4.1) cl = /CT, A = If, + a = si coshkili(z - zs), f’ = :tanhkili(z - $0)

4.2) cl = -kf, A = --If, + a = si coshki2i(z - zs), f’ = -:tanhki1i(z - zs)

4.3) cl = kf, A = -11, * a=slcosklzl(a:-~~), f’= +nklzl(z-20)

4.4) cl = -I$, A = If, + a = s1 cosklZl(z -x0), f’ = +anklll(s - x0)

4.5) cl = 0, A # 0, + a = klz, f’ = l/Ax

4.6) cl # 0, A = 0, + a = ICI, f’ = clz

Also we bypass the solutions of the velocity here considering the length of this article.

2 Axisymmetric cylindrical coordinate system

For the axisymmetric complex lamellar flow, because a/% = 0, we will have:

u -A?? f- ar’

wp = 0, Wg =

The steady Euler equation is reduced to13]:

Ug = 0,

dXdq axaq -- - -- aZaT dTdZ’

wz = 0

_. awe , _ aWe we%- 'u,,Tt'u,~= y

Also we let A = a(r)b(z), 77 = f(T) + g( ) z , so the incompressible condition becomes:

(5)

[y + (;+ ;)f/] + (f+ $) =o So we would get:

f” f (; + $-’ = C] b’

g” + Tg’ = -c1

(6)

20 Communications in Nonlinear Science & Numerical Simulation March 2000

On account of f’g’ # 0, Eq. (5) is simplified to:

a” +2flbl=o

r bg’

We discuss the following two situations:

1) If f’ = Kr, then we would have cl = 2K, a = A = const, b = Bz + C, and

g’=-K(Bz+C)+&

so the velocity is:

u = (A(B~ + C)K~, 0, -$$$B~ + c)” + AN)

Commonly, we let A = 1, C = 0, K = 1, N = 0, so the simplest form of the velocity is:

u = (Bzr, 0, -Bz”)

2) If b’ = Kbg’, we will obtain:

a’ I a” - - _ cl -

r ‘I + 2Kfl = m = const a T

g” + K(g’)2 = --cl, b” = (m - qK)b

We can get a special solution of (6) and (7), that is:

(7)

(8)

f’ = &r, a = A = const

For (B), we also have to analyze the sign of K, cl, and m - cl K, so it will be divided into many different situations to discuss, just as the situations in (3) and (4). It will not be difficult for us, but it is too tedious. So we also bypass these parts.

Some results of the complex lamellar flow will be given later.

Acknowledgment

This paper is supported by the National Basic Projects “Nonlinear Science” and “Frontier Problems in Fluid Mechanics and Aerodynamics”.

References

[l] Wu Jiezhi, Ma Huiyang and Zhou Mingde, Introduction to Vorticity and Vortex Dynamics, Adv. Education Press, Beijing, 1993 (in Chinese)

[2] Stirnan, P. G., Vortex Dynamics, Cambridge Univ. Press, Cambridge, 1992 [3] Batchelor, G. K., An Introduction to Fluid Dynamics, Cambridge Univ. Press, Cambridge, 1970