Basics of Fluid Mechanics, 2014a

560 CHAPTER 13. MULTI–PHASE FLOW
The velocity can vanish (zero) inside the film in another point which can be
obtained from
0= ρ ( )
L g x
2
μ L 2 − hx + τ i x
(13.65)
μ L
The solution for equation (13.65) is
x| @UL =0 =2h − 2 τ i
μ L gρ L
(13.66)
The maximum x value is limited by the liquid film thickness, h. The minimum shear
stress that start to create reversible velocity is obtained when x = h which is
0= ρ ( )
L g h
2
μ L 2 − hh + τ i h
(13.67)
μ L
↩→ τ i0 = hgρ L
2
If the shear stress is below this critical shear stress τ i0 then no part of the liquid will have
a reversed velocity. The notation of τ i0 denotes the special value at which a starting
shear stress value is obtained to have reversed flow. The point where the liquid flow
rate is zero is important and it is referred to as initial flashing point.
The flow rate can be calculated by integrating the velocity across the entire liquid
thickness of the film.
∫
Q h ∫ h
[ ( )
w = ρL g x
2
U y dx =
μ L 2 − hx + τ ]
i x
dx (13.68)
μ L
0
0
Where w is the thickness of the conduit (see Figure 13.15).
(13.68) results in
Integration equation
Q
w = h2 (3 τ i − 2 ghρ L )
6 μ L
(13.69)
It is interesting to find the point where the liquid mass flow rate is zero. This point can
be obtained when equation (13.69) is equated to zero. There are three solutions for
equation (13.69). The first two solutions are identical in which the film height is h =0
and the liquid flow rate is zero. But, also, the flow rate is zero when 3 τ i =2ghρ L .
This request is identical to the demand in which
τ icritical = 2 ghρ L
(13.70)
3
This critical shear stress, for a given film thickness, reduces the flow rate to zero or
effectively “drying” the liquid (which is different then equation (13.67)).