Question

Stokes' law is used to calculate the settling rate of barite in drilling fluid for two batches, P and Q. The particle size of batch P is 75 micrometers and the particle size of batch Q is 25 micrometers. The particle settling rate in the drilling fluid for batch P is ________ times that for batch Q.

• A. Four
• B. Nine
• C. Two
• D. One

Hint:

Stokes' law settling velocity is proportional to the square of particle diameter, so take the ratio of diameters and square it.

Answer 1

The Correct Option is B

Step 1: Write down Stokes' law for settling velocity:
Stokes' law gives the terminal settling velocity \(v\) of a small spherical particle falling through a viscous fluid under gravity as \[ v = \frac{d^{2}(\rho_{s}-\rho_{f})g}{18\mu} \] where \(d\) is the particle diameter, \(\rho_{s}\) is the particle (barite) density, \(\rho_{f}\) is the fluid density, \(g\) is gravitational acceleration and \(\mu\) is the fluid viscosity.
Step 2: Identify which quantities are the same for both batches:
Both batches P and Q consist of the same barite material settling in the same drilling fluid, so \(\rho_{s}\), \(\rho_{f}\), \(g\) and \(\mu\) are identical for both. The only quantity that differs between the two batches is the particle diameter \(d\), so the settling velocity is directly proportional to the square of the particle diameter, \(v \propto d^{2}\).
Step 3: Set up the ratio of settling velocities:
Using this proportionality, the ratio of the settling velocity of batch P to that of batch Q is \[ \frac{v_{P}}{v_{Q}} = \left(\frac{d_{P}}{d_{Q}}\right)^{2} \]
Step 4: Substitute the given particle sizes:
With \(d_{P} = 75\ \mu m\) and \(d_{Q} = 25\ \mu m\), \[ \frac{v_{P}}{v_{Q}} = \left(\frac{75}{25}\right)^{2} = (3)^{2} = 9 \] So the settling rate of batch P is nine times that of batch Q.
Final Answer:
\[ \boxed{\dfrac{v_{P}}{v_{Q}} = 9\ \text{(Option B, Nine)}} \]