Question

The terminal velocity of the fat globule in Stoke's law region is given by which of the following ?

• A. \( v_t = \frac{gD_p^2(\rho_p - \rho)}{18\mu} \)
• B. \( v_t = \frac{gD_p^2(\rho_p - \rho)}{24\mu} \)
• C. \( v_t = \frac{gR_p^2(\rho_p - \rho)}{18\mu} \)
• D. \( v_t = \frac{gR_p(\rho_p - \rho)}{18\mu} \)

Hint:

Always note the variable used in the denominator:
- If using radius (\( R_p \)), the denominator is \( 9 \).
- If using diameter (\( D_p \)), the denominator is \( 18 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Stoke's Law describes the settling or rising velocity of a spherical particle suspended in a viscous fluid.
In milk, fat globules are less dense than the continuous skim milk phase, causing them to rise (cream) over time under the influence of gravity.
Key Formula or Approach:
The standard expression of Stoke's Law in terms of particle radius (\( R_p \)) is:
\[ v_t = \frac{2 R_p^2 g (\rho_p - \rho)}{9 \mu} \]

Step 2: Detailed Explanation:
Let us express the equation in terms of the particle diameter (\( D_p \)).
Since \( D_p = 2 R_p \implies R_p = \frac{D_p}{2} \), we can substitute this into the equation:
\[ v_t = \frac{2 \left(\frac{D_p}{2}\right)^2 g (\rho_p - \rho)}{9 \mu} \]
\[ v_t = \frac{2 \frac{D_p^2}{4} g (\rho_p - \rho)}{9 \mu} \]
\[ v_t = \frac{D_p^2 g (\rho_p - \rho)}{2 \times 9 \mu} = \frac{g D_p^2 (\rho_p - \rho)}{18 \mu} \]
where:
\( v_t \) is the terminal velocity,
\( g \) is the acceleration due to gravity,
\( D_p \) is the diameter of the fat globule,
\( \rho_p \) is the density of the fat globule,
\( \rho \) is the density of the continuous liquid medium (plasma),
\( \mu \) is the dynamic viscosity of the medium.
This matches option 1.

Step 3: Final Answer:
The terminal velocity is given by \( v_t = \frac{gD_p^2(\rho_p - \rho)}{18\mu} \).