Page 351 - Modelling in Transport Phenomena A Conceptual Approach

9.1. MOMENTUM TRANSPORT 331

For the liquid phase the velocity components are simplified according to Figure
8.2. Since v, = v,(x) and vx = vy = 0, Table C.1 in Appendix C indicates that the
~
only non-zero shear-stress component is T ~ Hence, the components of the total
.
momentum flux are given by
L L LL L- L- L dv,L (9.1-37)
Tse=Tx%+(P V%)% -Txz---P
TL = 7y% + (PLVXL) .Y” = 0 (9.1-38)
L
yz
L
Tz, = T:, + (pLvf) v,” = pL ($) 2 (9.1-39)
The pressure, on the other hand, depends only on z. Therefore, only the
z-component of the equation of motion should be considered.
For a rectangular differential volume element of thickness Ax, length Az and
width W, as shown in Figure 9.2, Eq. (9.1-2) is expressed as
I,+&
( T2Z 1, Ax + Tk - ( Tt% Ax + n& [,+Ax
1,
+ (P”1, - PLIZ+,,) WAX + pLgWAxAz = 0 (9.1-40)
Dividing each term by W Ax Az and taking the limit as Ax + 0 and Az 4 0 gives

(9.1-42)
Substitution of Eqs. (9.1-37) and (9.1-39) into Eq. (9.1-42) and noting that
bv:/dz = 0 yields
L W dPL +pLg (9.1-43)
-P d22=-- dz
Now, it is necessary to write down the z-component of the equation of motion
for the stagnant air. Over a differential volume element of thickness Ax, length
Az and width W, Eq. (9.1-2) is written as
( pAlz - P~(++~=) WAX + pAg w aXaz = o (9.1-44)

Dividing each term by W Ax Az and taking the limit as AZ + 0 gives

lim pAlz - PAIz+At +PAg = 0 (9.1-45)
Ax-0 AZ

dPA A
-=p 9 (9.1-46)
dz