Jökull - 01.12.1966, Blaðsíða 39

TABLE II
Method due to: Number G Remarks
R. James (i) 47.4 Kg/sec Equation (1)
Isbin, Moy and Cruz (ii) 18.0 (8)
Homogeneous (iii) 26.5 (9)
Fauske (iv) 26.5 » (14)
Cruver (V) 41.9 Supersaturation Eqtn
35.8 Equilibrium (15)
Flow Separation 40.0-45.0 Apparatus,
(based on the measured mass flow of the liquicl fraction and bottom hole temperature data) Fig. 1
where E is the kinetic energy per unit mass
of the mixture, and e the efficiency of the
production of kinetic energy. Einally, (4) the
two phases are in thermal equilibrium, that is,
they have the same temperature which is a
function of the pressure only.
These assumptions require the following com-
ments. The assumption of a constant slip ratio
is no doubt incorrect. However, in most practi-
cal cases, where the dryness fraction is not
very low, the bulk of the kinetic energy is
carried by the vapour phase, and this assump-
tion therefore does not appear to introduce
gross errors. The validity of the polytropic
equation can be tested on the basis of steam-
table data. It can be shown that it is a fair
approximation. Alternatively, the polytropic
exponent n can be evaluated analytically as
shown in the Appendix. Moreover, the intro-
duction of the efficiency e in the third assump-
tion is made in order to emphasize the fact
that in the case of two-phase mixtures, the
conversion of enthalphy to kinetic energy
necessarily involves irreversible losses due to
frictional forces between the two phases. It is
obvious that e will increase with an increasing
dryness fraction and has the maximum value
e = 1. Einally, although the assumption of
thermal equilibrium appears rather reasonable,
it must be recognized that some disequilibrium
may occur at the relatively high velocities in-
volved in the critical flow of steam-water
mixtures. Thermal disequilibrium in an ex-
panding mixture lowers the expansivity and
therefore enhances the apparent polytropic
exponent n. It can possibly be detected on the
basis of experimental data on this figure.
It is now possible to proceed in the following
way.
(i) Continuity of mass flow requires that
xG = VgegRg (17)
(1 — x)G = V(ef(l — Rg) (18)
whence
1 G = X 1 —X vgeg vfef (19)
Since vg»v(, vm = xvg, vg=I/eg, (19) can be re-arranged to read equation
c. V*°" -
J ! (l-x)K0,n ef (20)
Also, G = Vgea, where ea is the density of the two-phase mixture apparent
9g
(1 — x)Keg x + — öf (21)
JÖKULL 193