88
Chapter 3
The Equations of Change for Isothermal Systems
From Bird, Stewart, Lightfoot
Transport Phenomena, Wiley
We now turn to the illustrative examples. The first two are problems that were discussed in the preceding chapter; we rework these just to illustrate the use of the equations of change. Then we consider some other problems that would be difficult to set up
by the shell balance method of Chapter 2.
EXAMPLE 3.6-1
Steady Flow in a Long
Circular Tube
Rework the tube-flow problem of Example 2.3-1 using the equations of continuity and motion. This illustrates the use of the tabulated equations for constant viscosity and density in
cylindrical coordinates, given in Appendix B.
SOLUTION
We postulate that v = bzvz{r, z). This postulate implies that there is no radial flow (vr = 0) and
no tangential flow (ve = 0), and that vz does not depend on 0. Consequently, we can discard
many terms from the tabulated equations of change, leaving
equation of continuity
(3.6-1)
r-equation of motion
(3.6-2)
0-equation of motion
0= -
дв
z-equation of motion
(3.6-3)
(3.6-4)
The first equation indicates that vz depends only on r; hence the partial derivatives in the second term on the right side of Eq. 3.6-4 can be replaced by ordinary derivatives. By using the
modified pressure & = p + pgh (where h is the height above some arbitrary datum plane), we
avoid the necessity of calculating the components of g in cylindrical coordinates, and we obtain a solution valid for any orientation of the axis of the tube.
Equations 3.6-2 and 3.6-3 show that <3> is a function of z alone, and the partial derivative
in the first term of Eq. 3.6-4 may be replaced by an ordinary derivative. The only way that we
can have a function of r plus a function of z equal to zero is for each term individually to be a
constant—say, Co—so that Eq. 3.6-4 reduces to
dv
(3.6-5)
The ty equation can be integrated at once. The ^-equation can be integrated by merely "peeling off" one operation after another on the left side (do not "work out" the compound derivative there). This gives
9 = Qz + C,
(3.6-6)
vz = ^ r2 + C2 In r + C3
(3.6-7)
The four constants of integration can be found from the boundary conditions:
B.C.I
B.C. 2
B.C3
B.C. 4
at z = 0,
at z = L,
at r = R,
at r = 0,
v7 = finite
(3.6-8)
(3.6-9)
(3.6-10)
(3.6-11)
The resulting solutions are:
(3.6-12)
2
v7
= •
Po - <3>L)R
4/JLL
(3.6-13)
§3.6
Use of the Equations of Change to Solve Flow Problems
89
Equation 3.6-13 is the same as Eq. 2.3-18. The pressure profile in Eq. 3.6-12 was not obtained
in Example 2.3-1, but was tacitly postulated; we could have done that here, too, but we chose
to work with a minimal number of postulates.
As pointed out in Example 2.3-1, Eq. 3.6-13 is valid only in the laminar-flow regime, and
at locations not too near the tube entrance and exit. For Reynolds numbers above about 2100,
a turbulent-flow regime exists downstream of the entrance region, and Eq. 3.6-13 is no longer
valid.
EXAMPLE 3.6-2
Falling Film with
Variable Viscosity
Set up the problem in Example 2.2-2 by using the equations of Appendix B. This illustrates
the use of the equation of motion in terms of т.
SOLUTION
As in Example 2.2-2 we postulate a steady-state flow with constant density, but with viscosity
depending on x. We postulate, as before, that the x- and y-components of the velocity are zero
and that vz = vz(x). With these postulates, the equation of continuity is identically satisfied.
According to Table B.I, the only nonzero components of т are rxz = TZX = —\x(dvjdx). The
components of the equation of motion in terms of т are, from Table B.5,
dp
O=~
+ pgsmp
dp
0= —f
др
(3.6-14)
(3.6-15)
л
(3.6-16)
where /3 is the angle shown in Fig. 2.2-2.
Integration of Eq. 3.6-14 gives
p = pgx sin j8
(3.6-17)
in which/(y, z) is an arbitrary function. Equation 3.6-15 shows that/cannot be a function of y.
We next recognize that the pressure in the gas phase is very nearly constant at the prevailing
atmospheric pressure paim. Therefore, at the gas-liquid interface x = 0, the pressure is also
constant at the value p a t m . Consequently, / can be set equal to p a t m and we obtain finally
p = pgx sin p + patn
(3.6-18)
Equation 3.5-16 then becomes
(3.6-19)
which is the same as Eq. 2.2-10. The remainder of the solution is the same as in §2.2.
EXAMPLE 3.6-3
Operation of a Couette
Viscometer
We mentioned earlier that the measurement of pressure difference vs. mass flow rate through
a cylindrical tube is the basis for the determination of viscosity in commercial capillary viscometers. The viscosity may also be determined by measuring the torque required to turn a
solid object in contact with a fluid. The forerunner of all rotational viscometers is the Couette
instrument, which is sketched in Fig. 3.6-1.
The fluid is placed in the cup, and the cup is then made to rotate with a constant angular
velocity ft0 (the subscript "o" stands for outer). The rotating viscous liquid causes the suspended bob to turn until the torque produced by the momentum transfer in the fluid equals
the product of the torsion constant k, and the angular displacement вь of the bob. The angular
displacement can be measured by observing the deflection of a light beam reflected from a
mirror mounted on the bob. The conditions of measurement are controlled so that there is a
steady, tangential, laminar flow in the annular region between the two coaxial cylinders
90
Chapter 3
The Equations of Change for Isothermal Systems
Fixed
Torsion wire with torsion constant kt
Outer cylinder
rotating /
v0 is a function of r
Bob is suspended
and free to rotate
Mirror
A,
Fixed
cylindrical
surfaces
In this У
region t h e \
fluid is
moving
with
ve = ve(r)
Inner cylinder
stationary
Rotating
cylindrical
cup
Fluid inside
is stationary
(b)
(a)
Fig. 3.6-1. (fl)Tangential laminar flow of an incompressible fluid in the space between two cylinders; the outer
one is moving with an angular velocity il0. (b) A diagram of a Couette viscometer. One measures the angular
velocity Cl0 of the cup and the deflection 6B of the bob at steady-state operation. Equation 3.6-31 gives the viscosity /л in terms of Д, and the torque Tz = kt6B.
shown in the figure. Because of the arrangement used, end effects over the region including
the bob height L are negligible.
To analyze this measurement, we apply the equations of continuity and motion for constant p and fi to the tangential flow in the annular region around the bob. Ultimately we want
an expression for the viscosity in terms of (the z-component of) the torque 7\ on the inner
cylinder, the angular velocity fl0 of the rotating cup, the bob height L, and the radii KR and R
of the bob and cup, respectively.
SOLUTION
In the portion of the annulus under consideration the fluid moves in a circular pattern. Reasonable postulates for the velocity and pressure are: v() = vn{r), vr = 0, V-. = 0, and p = p{r, z).
We expect p to depend on z because of gravity and on r because of the centrifugal force.
For these postulates all the terms in the equation of continuity are zero, and the components of the equation of motion simplify to
r-component
^-component
z-component
Щ
dp
~PT=~~a~
d
(3.6-20)
(3.6-21)
(3.6-22)
The second equation gives the velocity distribution. The third equation gives the effect of
gravity on the pressure (the hydrostatic effect), and the first equation tells how the centrifugal
force affects the pressure. For the problem at hand we need only the ^-component of the
equation of motion.2
2
See R. B. Bird, С F. Curtiss, and W. E. Stewart, Chem. Eng. Sci., 11,114-117 (1959) for a method of
getting p(r, z) for this system. The time-dependent buildup to the steady-state profiles is given by R. B.
Bird and С F. Curtiss, Chem. Eng. Sci., 11,108-113 (1959).
§3.6
Use of the Equations of Change to Solve Flow Problems
91
A novice might have a compelling urge to perform the differentiations in Eq. 3.6-21 before solving the differential equation, but this should not be done. All one has to do is "peel
off" one operation at a time—just the way you undress—as follows:
\ jr (rv0) = d
(3.6-23)
-f (rv0) = C}r
2
(3.6-24)
rvo = ^C,r + C2
(3.6-25)
ve = | C,r + ^
(3.6-26)
The boundary conditions are that the fluid does not slip at the two cylindrical surfaces:
B.C. 1
at r = KR,
B.C. 2
a t r = R,
vft = 0
(3.6-27)
vo = CLGR
(3.6-28)
These boundary conditions can be used to get the constants of integration, which are then inserted in Eq. 3.6-26. This gives
(
\KR
vtl = UOR —f
^
)
(3.6-29)
By writing the result in this form, with similar terms in the numerator and denominator, it is clear
that both boundary conditions are satisfied and that the equation is dimensionally consistent.
From the velocity distribution we can find the momentum flux by using Table B.2:
The torque acting on the inner cylinder is then given by the product of the inward momentum flux (~тгв), the surface of the cylinder, and the lever arm, as follows:
T, = (-rr0)\r-Kl<
• ITTKRL • KR = 4тгдЦ,Я 2 Ц — ^
\1 - кг)
(3.6-31)
The torque is also given by Tz = kfib. Therefore, measurement of the angular velocity of the
cup and the angular deflection of the bob makes it possible to determine the viscosity. The
same kind of analysis is available for other rotational viscometers.3
For any viscometer it is essential to know when turbulence will occur. The critical
Reynolds number (fi0R2p//x,)crit, above which the system becomes turbulent, is shown in Fig.
3.6-2 as a function of the radius ratio к.
One might ask what happens if we hold the outer cylinder fixed and cause the inner
cylinder to rotate with an angular velocity П, (the subscript "/" stands for inner). Then
the velocity distribution is
(R
f
(3.6-32)
This is obtained by making the same postulates (see before Eq. 3.6-20) and solving the
same differential equation (Eq. 3.6-21), but with a different set of boundary conditions.
3
J. R. VanWazer, J. W. Lyons, K. Y. Kim, and R. E. Colwell, Viscosity and Flow Measurement, Wiley,
New York (1963); K. Walters, Rheometry, Chapman and Hall, London (1975).