The flow of an incompressible fluid in a tube or other channel of constant cross section is said to be fully developed if the velocity does not vary in the of flow it is independent of axial position). This only beyond a certain distance from the tube inlet, called the entrance length. The entrance length depends on the tube diameter and Reynolds number, as discussed in Section 6.5. Fully developed flows are ordinarily indeed, they are the most important type of unidirectional Row. The gen- eral features of such Rows are discussed first, followed by specific examples.
Consider the flow of an incompressible fluid in a tube of arbitrary cross-sectional shape, as depicted in Fig. The axial coordinate is and the velocity field is assumed to be fully developed, so that v = t) only. It is to identify the conditions
which the flow will be unidirectional for which = Given that the continuity equation for this three-dimensional situation reduces to
examine the possibility of flow in the plane we define a stream function t) as flow in a tube of any cross-sectional shape will be unidirectional.
The Navier-Stokes equation written in of the dynamic pressure is
With = =0, all terms in the y and components of this equation vanish, indicating that = = or that = (x, t ) only. O f importance, the inertial in the x component also vanish. That is, for this fully developed flow. The compo- nent of (6.2-4) is reduced then to a linear partial differential equation, which for
flow is
Because t) and t), can depend at most on time. In other words, none of the other terms in (6.2-5) depend on so that cannot be a
of either. For steady, fully developed flow we obtain
Here 9 = only, so that must be a constant.
Equations (6.2-5) and (6.2-6) are the starting points for analyses of the fully devel- oped flow of Newtonian fluids. The simplifications in the Cauchy momentum equation for fully developed flow are Applications of Eq. (6.2-6) and the analogous mo- mentum equation for cylindrical coordinates are illustrated by the examples which follow.
Flow in a Parallel-Plate Channel The objective is to determine the steady, fully developed velocity field for pressure-driven flow in a channel with flat, parallel walls, shown in Fig. 6 2 . This is called plane It is assumed that the dimension is large.
so that = only; see Example 6.5-2 for a discussion of edge effects in channels of rectangu- lar cross section. Equation reduces to
d2 I (6.2-7)
Flow in a parallel-plate channel with spacing and mean velocity
with no-slip boundary conditions given by
The integration of is simplified by the fact that is a constant. Integrating twice and evaluating the two integration constants using (6.2-8) gives
The great advantage in using instead of P is that (6.2-9) is valid for a channel with
any spatial orientation. If the is 0), then if the channel
is vertical and there is no applied pressure 0), then Note too that the same parabolic velocity profile results if one of the no-slip boundary conditions is replaced with a symmetry condition at y = namely = 0 =
The mean velocity in the channel (U) is related to the pressure drop by
so that (6.2-9) may be rewritten as
It is seen that the maximum (center-plane) velocity The volumetric flow rate per unit width of channel is
Example Flow of a Power-Law Fluid in a Consider steady Row of a power-law fluid in a cylindrical tube of radius For this fully developed, flow we
= = and 9 = With such a velocity field, (or is the only component of the rate-of-deformation tensor. It follows from (5.6-3) that., for a Newton- ian generalized Newtonian fluid, (or %) is the only nonzero component of the viscous that only. From the component of the momentum equation we obtain
= (6.2-13) is satisfied if both sides a constant; that is, must be constant. Integrating, it is found that
Steady Flow with a Pressure Gradient 255
where C, is a constant. A symmetry argument like that used to derive (2.2-22) leads to the conclusion that = at r = that and
The fact that there will be a pressure drop in the direction of flow indicates that For a power-law fluid it is found from (5.6-13) and Table 5-10 that
The minus sign in (6.2-1 6) is required because 0, and is used because Combining (6.2-1 5) and (6.2-1 6) and solving for gives
The solution is completed by integrating (6.2-17) and using the no-slip condition, =0, to evaluate the integration constant. The result is
where is the mean velocity. The shape of the velocity profile in (6.2-18) is determined by the power-law exponent, n. For the velocity approaches the mean velocity.
approximating plug flow. (As indicated in Chapter 5, polymeric fluids exhibit values of n as small as 0.2.) In general, the volumetric flow rate is given by
For a Newtonian fluid, where n = 1 and = (6.2-18)-(6.2-20) yield the well-known results for flow,
d z '
Thus, the velocity profile for a Newtonian fluid in a circular is parabolic, as for a plate channel, but the maximum (centerline) velocity in case is 2U. Equation is one form of Poiseuille law.
'This type of flow is named after Jean Poiseuille a French physician whose interest in of blood flow led him conduct an extensive set of and flow measurements in Paris in the 1830s and He employed water and other liquids in glass capillary tubes. and was extremely
Example Flow of Immiscible Fluids in a Parallel-Plate Channel A simple type of two-phase flow occurs when immiscible fluids occupy distinct layers in a parallel-plate channel, as depicted in Fig. The density and viscosity of fluid and may differ from those of 2 and It desired to determine the steady, fully developed velocities of the two fluids, which are denoted as and respectively,
The equation for each phase reduces to
Integrating this twice gives
where and are constants. These four constants are determined by the conditions
Equations (6.2-26) and (6.2-27) are the usual no-slip conditions at the solid surfaces, whereas (6.2-28) and (6.2-29) express the matching of the tangential components of velocity and stress at the fluid-fluid interface. It is convenient to introduce the constants
where has units of velocity [compare with Eq. and is dimensionless. The velocities in the two fluids are written as
Fluid 2
4
Fluid 1
Figure 6 3 . Flow of immiscible fluids in a channel.
the same time, dependence of Q on was revealed also by a less extensive and precise made in Berlin a German engineer,
called equation. The first derivation of this result the
attributed to the Hagenbach (1833-1910) of in 1860, although several to have done this at about the time. Incidentally, invented the manometer, to measure arterial pressures in animals and
Steady Flow with a Pressure Gradient 257
where and are assumed to be known.
What remains is to relate the pressures in the two fluids. Examples 6.2-1 and 6.2-2 it was seen that for fully developed flow of a single, incompressible fluid in a channel of known dimensions, specifying the mean velocity was the same as setting the axial gradient of the dy- namic pressure [see Eqs. (6.2-lo), and Extra care is needed in the present problem, because while constant within each fluid, is generally not the same in two phases. The constraints on the two-phase flow are revealed by considering the actual pressure, For the general case of a channel inclined at an arbitrary angle, y, even in the absence of Row, because of static pressure variations. However, from the definition of the dynamic pressure, Eq. it follows that for this flow
Thus, the constancy of implies that too is constant within each phase. The final piece of information needed comes from the normal stress balance at the fluid-fluid interface, based on (5.7-9). Given that the interface is flat and that there are no viscous stresses the values of P there must match. We conclude has the same constant value throughout both fluids. To emphasize that there is only one independent pressure gradient,
. . (6.2-30) is rewritten as
The need to consider actual pressure, and not just dynamic pressure, is typical of problems involv- ing fluid-fluid interfaces.
Example 6.2-4 Flow of a Liquid Down a n Inclined Surface With reference to Fig. 6-4, the objective is to determine the velocity of a liquid film flowing down a surface which is oriented at an angle relative to vertical. The film thickness (H) is assumed to be constant, making it possible to have steady, fully developed flow. Thus, it is assumed that only and
= = from which it follows (as before) that 9 = only and that is constant.
The falling liquid film can be viewed as a special case of the two-fluid problem in Example 6.2-3, in which fluid 1 is now the liquid and 2 the gas. Assuming that the gas occupies a
Figure 6-4. Flow of a liquid film down an inclined face.
Flow with a Moving Surface 259 space at least as thick as the liquid film, and that a typical ratio of liquid to gas viscosit-
ies is
-
(Chapter the parameter by (6.2-31) will be extremely large. Assum- ing also that the gas pressure is the reasoning leading to (6.2-35) indicates thatin the liquid. Noting that = cos it follows from (6.2-32) and (6.2-35) that the
liquid is
Fluid
.
where the subscript L denotes liquid properties. The velocity profile can be rewritten in of the mean velocity as
Figure Couette viscometer. The outer cylinder is rotated at angular velocity and the inner cylinder is fixed. (a) Overall view; (b) enlargement of area indicated by dashed rectangle in (a) for with and =
cos
Notice that (6.2-37) is exactly the same as the result obtained in Example 6.2-1 for flow in a channel of half-width H.
Assuming that as done in deriving is the same as neglecting the shear stress exerted by the gas on the liquid. As discussed in Section 5.7, this is a common approxima- tion at gas-liquid interfaces, and it has the effect of making the liquid velocity independent of the gas properties. It is readily confirmed that (6.2-36) is obtained also by solving
fluid-filled gap is of thickness and the linear velocity at the wetted surface of the outer cylinder is Noting the circular symmetry and neglecting end effects, it is assumed that v and 9 do not depend on or Similar to fully developed flow in a channel, a velocity field of the form
and = = is consistent with continuity.
From the component of the Navier-Stokes equation it follows that
, Integrating (6.3-1) twice yields terms in involving and Evaluating the integration constants using the no-slip boundary conditions,
with the boundary conditions
we obtain This is clearly the preferred approach if one is interested only in the liquid, in it avoids
having to determine the velocity in the gas.
6.3 STEADY FLOW WITH
A SURFACE . ,.
The component of momentum serves only to determine and therefore it provides
, no additional information concerning the velocity. The component only confirms that for the
-A- assumed of v, must be. independent of
, Couette viscometers are normally designed with gaps, which leads to a simpler velocity profile. For 1 the effects of surface curvature will be negligible, as in the first approx- imation to the concentration field in Example Reformulating the problem using local rectan- gular coordinates, as shown in Fig. conservation of momentum gives
is induced not just by a gradient in dynamic pressure, but also by motion at a solid or fluid interface. Flows caused by the tangential movement of a surface provide purest illustrations of surface-driven flow, in that there need not be any pressure gradient in the flow direction. two examples in this section, both of which entail of a solid surface, in that category. The main feature that distinguishes these rotational flows from the other unidirectional flows in this chapter is that the
curved, implying that all material points are undergoing spatial accelerations. A pres- gradient normal to the direction of flow is needed to sustain such
accelerations. The velocity profile for small gaps is then simply
Flow flow in the annular gap cylinders,
of which is rotated at a constant angular velocity, is termed In a Couette as shown in Fig. the inner cylinder is fixed and the outer one is rotated. The
, .
where and flow for is referred
to as plane An identical, linear velocity profile is obtained (more laboriously)
expanding the numerator and denominator of (6.3-3) in Taylor series about = Local rectan- gular coordinates find extensive use in lubrication theory (Section 6.6) and boundary layer theory (Chapter 8). As seen here, they are appropriate when the smallest radius of curvature of the bounding greatly exceeds the other length scales governing the flow.
Measurements of the torque required to the outer cylinder (or to keep the inner cylinder stationary) provide an effective method to determine the viscosity of the fluid. The magnitude of the torque exerted on the fluid by the outer cylinder is where L is the wetted length of the cylinders. The shear stress, (Table is determined in general by differentiating Eq. (6.3-3). For narrow gaps, however, = computed from Eq.
(6.3-5). An important property of plane Couette is that the shear stress is independent of position, or = Using = = the torque for 1 is
Example Surface of a Liquid in Rigid-Body Rotation Figure 6-6 shows an example of a system with a gas-liquid interface of unknown shape, consisting of a liquid in an open container of radius R that is rotated at an angular velocity If the container is rotated long enough, a steady state is reached in which the liquid is in rigid-body rotation. It is desired to determine the steady-state interface height, assuming that the ambient air is at a constant pressure, Because the viscous vanishes for rigid-body rotation, the analysis apply to any liquid, Newtonian or non-Newtonian. The effects of surface tension will be neglected.
The velocity of a fluid in rigid-body rotation is the same as that in a rotating solid, or
= or (6.3-7)
with = Because of the free surface, P will be used as the pressure variable instead of The circular symmetry indicates that P = z) only; the component of the Cauchy momentum equation (Table 5-2) confirms this. Given that and -g, the and components of conservation of momentum become
I I
I Liquid
Figure 6-6. A liquid in rigid-body rotation in an open con- tainer.
Dependent Flow 261
Using Eq. (6.3-7) in (6.3-8) and integrating, the pressure is found to be of the form
where is an unknown function. Integrating Eq. another expression for the pressure is
where is an unknown function. The two unknown functions are identified by comparing Eqs.
and leading to the conclusion that
where c is an unknown constant.
To determine the height of the interface, we note that with viscous stresses absent and surface tension assumed to be negligible, the normal stress balance requires that the liquid pres- sure at the interface equal Setting = and P = in Eq. (6.3- 12) gives
Thus, the interface is parabolic in shape, with the lowest point at the center. The additional
, needed to determine is a specification of the total volume of fluid in the container, or dr.
Using (6.3-13) in (6.3-14) to calculate c, the final result is
where the liquid height under static conditions.