Miscible Treatment¶

A serious problem in equation of state modelling is to establish a phase identity-independent $$K_{rh}(S_h,S_w)$$ model which becomes a straight line when we approach the critical point. Otherwise the problem is unsolvable in this region as $$\partial(sat)/\partial(zmf) \Rightarrow\infty$$ at the critical.

A further problem, even if the system does not approach the critical point, is that, in a miscible flood (such as when CO2 invades hydrocarbon regions) the system phase envelope can move from right to left, so that a system can change from a liquid to a gas, without going through an intermediate two-phase state. This can produce a serious discontinuity which renders solving the non-linear equations very difficult:

The solution is to make the relative permeability and capillary pressure functions dependent on surface tension – which is reasonable, as we would expect zero capillary pressures and straight-line relative permeabilities in the case of zero interphase surface tension. Then we make the surface tension a function of the compositions of the phases. To do this we use a classical correlation for surface tension, the McLeod-Sugden expression [Sug24], which is a function of the component parachors, which become a new required set of input values.

The surface tension, $$\sigma$$, computed by [Sug24], such that $$\sigma \Rightarrow 0$$ as the oil and gas phase compositions converge, is used to define a miscibility fraction:

$f_m=1-f_i, \; f_i= \left(\frac{\sigma}{\sigma_{ref}} \right)^P$

Where $$\sigma_{ref}$$ is the surface tension in some immiscible reference state at which the rel. perms are measured. P is a parameter that goes from 0.15 to 0.25 (taken equal to 0.25).

While an ‘oil’ fraction can be defined as:

$f_o=\frac{T_c}{(T_c+T_{res})}$

Where $$T_c$$ and $$T_{res}$$ are the critical and reservoir temperatures, respectively. So that $$T_c=T_{res}$$, $$f_o=0.5$$; if $$T_c>>T_{res}$$ (critical above res temp), then $$f_o \Rightarrow 1$$.

The gas fraction is simply taken as the complement of the oil fraction:

$f_g=1-f_o$

With $$f_o$$ and $$f_g$$, the rel. perm of each of the hydrocarbon phases is expressed as follows:

$K_{rh} = f_o \, K_{ro} + f_g \, K_{rg}$

Where

$K_{rg}(S_g,S_w) = \frac{(S_o \, K_{rgo}(S_g) + S_w \, K_{rgw}(S_w))}{(S_o+S_w)}$

Four tables are required: $$K_{row}$$, $$K_{rog}$$, $$K _{rgo}$$ and $$K_{rgw}$$. If these are not available, $$K_{row}$$ is taken as $$K_{rgw}$$.

$$K_{rog}$$ and $$K_{rgo}$$ are computed as:

$K_{rog} = f_i K_{rog} (\mbox{scaled table val}) + f_m \, K_{rog} (\mbox{scaled straight line value}).$

Finally, the cap press pressure goes like $$P_{cog} \sim f_i$$ and vanishes with surface tension:

$P_{cog}(\mbox{scaled}) = f_i \, P_{cog}(\mbox{table})$

So, as $$f_m \Rightarrow 1$$, $$f_i \Rightarrow 0$$, criticals and cap pressure vanish.

The result is a much smoother system in the near-critical and miscible region. The hydrocarbon phase has properties which are functions of its composition, not its phase label.