Page 227 - Bird R.B. Transport phenomena

§7.6 Use of the Macroscopic Balances for Steady-State Problems 211

Fig. 7.6-2. Flow in a liquid-liq-
v 2
uid ejector pump.

:v 0
Plane 1 \ Stream la Plane 2
Stream lb

turbulent and the velocity profiles at planes 1 and 2 are assumed to be flat. In the following
analysis F^ s is neglected, since it is felt to be less important than the other terms in the mo-
mentum balance.
SOLUTION (a) Mass balance. At steady state, Eq. (A) of Table 7.6-1 gives

Ща + ЩЬ = W 2 (7.6-9)
or
pv (lS,) + p(^oXiSi) = pv S 2 (7.6-10)
2
o
Hence, since S] = S , this equation gives
2
= ~ 3 v (7.6-11)
v 2
for the velocity of the exit stream. We also note, for later use, that и? = w xb = \w .
2
1я
(b) Momentum balance. From Eq. (B) of Table 7.6-1 the component of the momentum bal-
ance in the flow direction is
+ • ,) - + p 2 S 2 ) = 0 (7.6-12)
(v u w u v xb w xb (v 2 w 2
or using the relation at the end of (a)

= &Щ + \v ) - lv )(p(iv )S ) (7.6-13)
2
o
o
0
from which
v
Pi ~ Pi = XsP o (7.6-14)
This is the expression for the pressure rise resulting from the mixing of the two streams.
(c) Angular momentum balance. This balance is not needed.
(d) Mechanical energy balance. Equation (D) of Table 7.6-1 gives

, = E v (7.6-15)

or, using the relation at the end of (a), we get
= E V (7.6-16)
Hence

v (7.6-17)
Wi 144 °
is the energy dissipation per unit mass. The preceding analysis gives fairly good results for
liquid-liquid ejector pumps. In gas-gas ejectors, however, the density varies significantly and
it is necessary to include the macroscopic total energy balance as well as an equation of state
in the analysis. This is discussed in Example 15.3-2.