Question

Consider a micellar displacement process in a homogeneous reservoir with a porosity of 30%. The volume of the microemulsion slug to be injected is 4% of the pore volume. The slug contains 4 vol% surfactant. The density of the rock and the surfactant is $2.7~\text{g/cm}^3$ and $1.1~\text{g/cm}^3$, respectively.

Assuming that the average surfactant adsorption is 0.25 mg/g of the reservoir rock, the fraction of the injected surfactant that will be adsorbed is ............. (rounded to two decimal places).

Hint:

Work with an arbitrary bulk volume $V$ so it cancels. Convert mg/g to g/g, then compute: fraction $=\dfrac{0.25\text{ mg/g}\times \rho_r(1-\phi)}{\rho_s\times (0.04\text{ slug})\times (0.04\text{ surfactant})\times \phi}$.

Answer 1

- Let bulk reservoir volume be $V$. Pore volume $= \phi V = 0.30V$.
- Slug volume $=0.04(\phi V)=0.04\times0.30V=0.012V$.
- Surfactant volume in slug $=0.04 \times 0.012V=0.00048V$.
- Injected surfactant mass: $M_{\text{inj}}=\rho_s(0.00048V)=1.1\times0.00048V=0.000528V$ g.
- Rock mass: $M_r=\rho_r(1-\phi)V=2.7\times0.70V=1.89V$ g.
- Adsorbed surfactant mass: $M_{\text{ads}}=0.25\,\text{mg/g}\times M_r=0.00025\times1.89V=0.0004725V$ g.
- Fraction adsorbed: \[ \frac{M_{\text{ads}}}{M_{\text{inj}}} =\frac{0.0004725V}{0.000528V}=0.8949\approx 0.89. \]