Page 59 - Bird R.B. Transport phenomena

44 Chapter 2 Shell Momentum Balances and Velocity Distributions in Laminar Flow

When this equation is divided by LW Ax, and the limit taken as Ax approaches zero, we
get

^
• ' " ° ~ ^ z z l z = L = pg cos p (2.2-7)
The first term on the left side is exactly the definition of the derivative of ф with respect
Х1
to x. Therefore Eq. 2.2-7 becomes
г-Ъ-фгАг-L ^
= p g p ( 2 2 g )
dx L
At this point we have to write out explicitly what the components ф and ф are, mak-
хг
2г
ing use of the definition of ф in Eqs. 1.7-1 to 3 and the expressions for r xz and r zz in Ap-
pendix B.I. This ensures that we do not miss out on any of the forms of momentum
transport. Hence we get
v v
Фхг = T xz + P x z = ~t*> y 1 + pv v (2.2-9a)
x z
ф 2г — p H~ T , + pv v = p — 2fji —^ + pv v (2.2-9b)
Z
z 7
z z
In accordance with the postulates that v z = v (x), v x = 0, v y = 0, and p = p(x), we see that
z
(i) since v x = 0, the pv v term in Eq. 2.2-9a is zero; (ii) since v z = v (x), the term
z
x z
-2l±(d /dz) in Eq. 2.2-9b is zero; (iii) since v z = v (x), the term pv v is the same at z = 0
z
z
z z
and z = L; and (iv) since p = p(x), the contribution p is the same at z = 0 and z = L. Hence
T XZ depends only on x, and Eq. 2.2-8 simplifies to
(2.2-10)
This is the differential equation for the momentum flux T . It may be integrated to give
XZ
r xz = (pg cos p)x + C, (2.2-11)
The constant of integration may be evaluated by using the boundary condition at the
gas-liquid interface (see §2.1):
B.C.I: atx = 0, r xz = 0 (2.2-12)
Substitution of this boundary condition into Eq. 2.2-11 shows that C } = 0. Therefore the
momentum-flux distribution is
r = (pg cos P)x (2.2-13)
xz
as shown in Fig. 2.2-3.
Next we substitute Newton's law of viscosity
*«=-"£ a2 i4)
-

into the left side of Eq. 2.2-13 to obtain
dv^ Jpg cos p
dx \ V-

which is the differential equation for the velocity distribution. It can be integrated to
give
/oe cos B
2 У + С 2 (2.2-16)