O/W Emulsion Creaming Velocity Calculation

Stokes' Law Formula

Creaming velocity ($v$) in an emulsion is calculated using Stokes' Law, which describes the terminal velocity of a particle in a fluid:

$ v = \frac{d^2 (\rho_c - \rho_d) g}{18 \eta} $

Where:

• $v$ = Velocity of creaming (cm/sec)
• $d$ = Average diameter of dispersed particles (cm)
• $\rho_c$ = Density of the continuous phase (g/cm$^3$)
• $\rho_d$ = Density of the dispersed phase (g/cm$^3$)
• $g$ = Acceleration due to gravity (cm/sec$^2$)
• $\eta$ = Viscosity of the continuous phase (poise)

Note: Since the dispersed phase (oil) is less dense than the continuous phase (water), it will rise, causing creaming. The formula uses the density difference $(\rho_c - \rho_d)$ to calculate the upward velocity.

Given Parameters & Unit Conversion

Parameters provided:

• Specific gravity of mineral oil (dispersed phase) = 0.9
• Specific gravity of aqueous phase (continuous phase) = 1.05
• Average particle diameter ($d$) = 5 micrometers ($\mu$m)
• Viscosity of external phase ($\eta$) = 0.5 poise
• Gravity constant ($g$) = $981 \text{ cm/sec}^2$

Convert units to cgs system (cm, g, sec):

• Density of oil ($\rho_d$) = $0.9 \times 1 \text{ g/cm}^3 = 0.9 \text{ g/cm}^3$
• Density of aqueous phase ($\rho_c$) = $1.05 \times 1 \text{ g/cm}^3 = 1.05 \text{ g/cm}^3$
• Particle diameter ($d$) = $5 \text{ µm} = 5 \times 10^{-4} \text{ cm}$

Creaming Velocity Calculation

Calculate the density difference:

$ \rho_c - \rho_d = 1.05 \text{ g/cm}^3 - 0.9 \text{ g/cm}^3 = 0.15 \text{ g/cm}^3 $

Substitute the values into Stokes' Law:

$ v = \frac{(5 \times 10^{-4} \text{ cm})^2 \times (0.15 \text{ g/cm}^3) \times (981 \text{ cm/sec}^2)}{18 \times (0.5 \text{ poise})} $

$ v = \frac{(25 \times 10^{-8} \text{ cm}^2) \times (0.15 \text{ g/cm}^3) \times (981 \text{ cm/sec}^2)}{9 \text{ g/(cm sec)}} $

$ v = \frac{3678.75 \times 10^{-8}}{9} \text{ cm/sec} $

$ v \approx 408.75 \times 10^{-8} \text{ cm/sec} $

$ v \approx 4.0875 \times 10^{-6} \text{ cm/sec} $

Unit Conversion to cm/day

Convert the velocity from cm/sec to cm/day:

$ 1 \text{ day} = 24 \text{ hours/day} \times 60 \text{ min/hour} \times 60 \text{ sec/min} = 86400 \text{ sec/day} $

$ v (\text{cm/day}) = (4.0875 \times 10^{-6} \text{ cm/sec}) \times (86400 \text{ sec/day}) $

$ v (\text{cm/day}) \approx 0.35316 \text{ cm/day} $

Rounding to two decimal places, the creaming velocity is approximately 0.35 cm/day.