Mathematical Proceedings of the Cambridge Philosophical Society Volume 69 issue 3 1971 [doi 10.1017_s0305004100046880] Packham_ B. A._ Shall_ R. -- Stratified laminar flow of two immiscible fluids.pdf
Mathematical Proceedings of the Cambridge Philosophical Society Volume 69 issue 3 1971 [doi 10.1017_s0305004100046880] Packham_ B. A._ Shall_ R. -- Stratified laminar flow of two immiscible fluids.pdf
Mathematical Proceedings of the Cambridge Philosophical Society Volume 69 issue 3 1971 [doi 10.1017_s0305004100046880] Packham_ B. A._ Shall_ R. -- Stratified laminar flow of two immiscible fluids.pdf
Abstract. We consider the steady co-current flow of two immiscible viscous liquids
in a horizontal pipe, the fluid interface being ripple-free and plane. It is shown that if
the cross-section of the duct is symmetric with respect to the interface, the velocity
distribution may be expressed in terms of two separate pipe-flow solutions. One
corresponds to the flow of a single fluid occupying the whole of the pipe, and the
second to a similar flow in a pipe whose cross-section coincides with that of the region
occupied by one fluid in the two-phase motion. Flow rates and other quantities of
engineering interest are evaluated, and several particular flows are discussed.
1. Introduction. The problem of the steady laminar flow of a single viscous liquid
in a straight pipe of constant cross-section is classical and solutions have been found
for a large number of cross-sections. There appears, however, to have been little
interest in the study of the stratified laminar flow of two immiscible liquids in hori-
zontal pipes until quite recently. This recent interest stems from the possibility of
reducing the power required to pump oil in a pipeline by the suitable addition of water.
Laminar flows with smooth horizontal ripple free interfaces have been obtained
experimentally by Charles and Lilleleht(3), who also give the analytic solution for a
rectangular pipe with horizontal and vertical sides when the ratio of the depths of the
two fluids is arbitrary. Using an integral equation method Dumitrescu and Staniscu(5)
had earlier obtained the solution for the special case of equal depths and also for a
circular pipe when the interface is a diametral plane. The latter problem has been
generalized by Bentwich(2), who treats the case of a circular fluid interface.
Numerical solutions for the circular pipe problem with a general position of the
horizontal interface have been given by Charles and Redberger(4), Gemmell and
Epstein (6) and Yu and Sparrow (11), and in the case when the cross-section is a segment
of a circle by Achutaramayya and Sleicher(l).
The problem for a general cross-section and general position of a plane interface is
of considerable difficulty. However, we will show that, when the cross-section is
symmetrical with respect to the interface, the solution may be easily expressed in
terms of the two separate solutions corresponding to the flow of a single liquid in the
whole pipe and in a half-pipe bounded by the interface. When each of these has a
known solution the required flow may be written down immediately.
2. The basic solution and flow parameters. Consider a steady pressure-driven two-
phase flow in a straight horizontal pipe whose bounding cross-section C is symmetric
444 B. A. PACKHAM AND R. SHAIL
with respect to the fluid interface. A system of Cartesian coordinates is used in which
the z-axis is parallel to the pipe wall and the fluid interface lies in the horizontal
(x, z) plane. The region of the duct for which y > 0 is occupied by a fluid of viscosity /tlt
whereas a fluid of viscosity fi2 occupies the lower portion y < 0.
We denote by ux[x, y) and u2(x, y) the axial fluid velocities in the regions y > 0 and
y < 0 respectively. Further, w^x, y) (i = 1,2) is defined to be the axial velocity
distribution of a fluid of viscosity /i{ which completely fills the pipe, the driving
pressure gradient P being the same as that in the two-phase flow. Thus ut and wi both
satisfy the equation
Conditions (ii) and (iii) express the continuity of velocity and shear stress at the
interface L.
We next introduce functions r{r{x, y) and W(x, y) which satisfy the equation
A I A g /n\
&e 2 &/ 2
and the boundary conditions
f{x,y) = Q on C, )
K>
Y(x,y) = 0 on C^UJLJ
It follows from symmetry that
i/r(x, -y) = f(x,y)
and Y(x, —y) = 0 on <72 U L.
p
Further wi = -— ^(x, y),
and the function (P/2^) Y(a;, ?/) gives the velocity distribution for a fluid of viscosity
ft1} flowing in a pipe whose boundary is Cx U L.
To construct a solution of the two-phase flow problem we represent ut as
and
Stratified laminar flow 445
i.e. the same as that of a fluid of viscosity i(/ti + ^ 2 ) which fills the pipe.
The flow rates Qi (i = 1,2) of the twofluidsare
* -l
where S1 denotes the portion y > 0 of S, the duct cross-section. Thus, the total flow
rate Q = Qx + Q2 is given by
- - IL \ Ii Jr I I
Cj/ ^— *; rII IT \ X * If) CbOCCuU ~j~ II llfiSCy If I CbOC Gulf • 1 1 1 )
along a pipe bounded by Cx U L, and the second term is the flow rate of afluidof
viscosity ^(ji1+/i2) which fills the whole pipe.
In discussing the effect of stratification on the volumetric flow rates of the separate
components several related physical parameters are commonly employed. One such
parameter is the volumetricflowrate factor Vi = Qi/Qi tun, where Qt tun is the flow rate
when the liquid * fills the whole pipe and the pressure gradient is the same as in the
two-phase system. The practical interest hi the stratification problem lies in the fact
that under suitable circumstances Yi may exceed unity.
Since p /• /•
Qttui\= «j—JJ f(x,y)dxdy, (12)
it follows that in the symmetric case considered above
, (13)
Q I to )l\ /HP*
We note that for the present problem all three parameters depend solely on R and
the ratio of the viscosities, and that R depends only on the geometry of the cross-section.
3. Some particular solutions. As remarked earlier, solutions for the symmetric two-
phase system may be written down immediately when the single liquid pipe flows are
known for both the half-pipe bounded by Cx U L and the full pipe bounded by C1U C2-
Many of these solutions are perhaps better known in the context of the St Venant
torsion problem of elastostatics (Timoshenko and Goodier(lO), Chapter 11). In this
context \F and i/r are the stress functions for the cylindrical rods whose cross-sections
are those of the half and full pipes respectively, and the integrals in equation (14)
are simply related to the torsional rigidities of the rods.
Consider first two-phaseflowin a circular or sectoral pipe. A general solution for \jr for
a sector of a circle was obtained by Greenhill(7) in the form of an infinite series. When
the angle 2a of the sector is mn/n (m and n integers) the series may be summed to give
a closed form solution. With the aid of these solutions, and an appropriate rotation of
axes, the solution of the corresponding two-phase problem follows immediately.
For example, for a circular pipe of unit radius we need GreenhiU's solution for
a = \TT, namely
o ^ ^ f , (20)
where kn = (2n +1) 77-/2a, is well known.
The solution W(x, y) for the half-pipe is, of course, obtained by replacing b by \b and
y by y — \b in equation (20).
From equation (9) the interfacial velocity in the symmetric case is thus given by
^ ^ (21)
n=o akl cosh&n&j
which agrees with that obtained by Dumitrescu and Staniscu(5) and Charles and
Lilleleht(3).
From p. 278 of Timoshenko and Goodier
» 1 {2n+\)-nb\
(22)
^ ^
Convergence is rapid for b > a (for b < a we interchange a and b), and with A = bja,
approximate formulae are
„ 1/A-
, „ l/l-0-
l/l-0-315A\
and **
*88 (( ll -006-3 0 * )
The following table of values of R for various A is easily calculated by using
equation (22):
A 00 10 5 4 3 2 ] 0-5 0-2 01 0
R 0-500 0-466 0-428 0-407 0-373 0-307 0-204 0 1 5 3 0-134 0-129 0125
An interesting variant of the above is afforded by a square pipe having one diagonal
horizontal. For a square pipe of side 2a with a side horizontal, \jr may be derived from
(20) by putting b = a. The function Y for the right-angled isosceles triangle bounded
by x = a, y = a and x + y = 0 was obtained by Kolossof (8), and is given by
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