Question

An oil droplet is to be mobilized by injecting water through a pore throat. The oil–water interface has the rear radius of curvature $r_A = 25\times10^{-6}\ \text{m}$ and the forward radius of curvature $r_B = 5\times10^{-6}\ \text{m}$. The pore is completely water-wet (contact angle $=0^\circ$) and interfacial tension is $\sigma = 0.025\ \text{N/m}$. The minimum pressure drop required to mobilize the trapped oil droplet is ________ N/m$^2$ (nearest integer). 

Hint:

To free a trapped droplet, overcome the {difference} in capillary pressures: $\Delta P_{\min}=2\sigma(1/r_{\text{front}}-1/r_{\text{rear}})$ for a water-wet system.

Answer 1

- For a water-wet capillary, capillary pressure: $P_c = \dfrac{2\sigma}{r}$ (since $\cos\theta=1$).
- Minimum driving $\Delta P$ equals the difference between forward and rear capillary pressures:
$\Delta P_{\min} = 2\sigma\!\left(\dfrac{1}{r_B}-\dfrac{1}{r_A}\right)$.
$\Rightarrow \Delta P_{\min}=2(0.025)\!\left(\dfrac{1}{5\times10^{-6}}-\dfrac{1}{25\times10^{-6}}\right)=8000\ \text{N/m}^2$.