๐ Understanding Stokes' Law
Stokes' Law describes the drag force experienced by a small spherical object moving through a viscous fluid. This law is crucial for understanding the motion of objects in fluids where viscous forces are significant.
๐ History and Background
Sir George Gabriel Stokes, an Irish physicist and mathematician, derived Stokes' Law in 1851. His work was fundamental in fluid dynamics and helped explain phenomena like the settling of particles in suspensions.
๐ Key Principles of Stokes' Law
- ๐ง Viscosity: ๐ก๏ธ The fluid's resistance to flow. Higher viscosity means greater resistance.
- โช Spherical Object: ๐ต The object must be spherical for the law to apply accurately.
- ๐ Low Reynolds Number: ๐ The flow around the object should be laminar (smooth), characterized by a low Reynolds number (typically less than 0.1).
๐งฎ The Formula
Stokes' Law is mathematically expressed as:
$F_d = 6 \pi \eta r v$
Where:
- ๐ช $F_d$ is the drag force.
- ๐ง $\eta$ (eta) is the dynamic viscosity of the fluid.
- ๐ด $r$ is the radius of the spherical object.
- ๐ $v$ is the velocity of the object relative to the fluid.
๐ Real-World Examples
- ๐งช Sedimentation: โณ Determining the settling rate of particles in a liquid, crucial in environmental science and chemical engineering.
- ๐ฉธ Blood Flow: โค๏ธ Modeling the movement of red blood cells in blood vessels.
- ๐ง๏ธ Raindrops: โ Understanding the terminal velocity of small raindrops in the atmosphere.
- ๐ญ Industrial Processes: โ๏ธ Designing filters and separation techniques in various industries.
โ๏ธ Solved Problems
Problem 1: Determining Drag Force
A spherical particle with a radius of 0.5 mm is falling through water with a viscosity of 0.001 Paยทs at a velocity of 0.1 m/s. Calculate the drag force.
Solution:
Using Stokes' Law:
$F_d = 6 \pi \eta r v$
Given:
- ๐ $r = 0.5 \text{ mm} = 0.0005 \text{ m}$
- ๐ง $\eta = 0.001 \text{ Pa} \cdot \text{s}$
- ๐ $v = 0.1 \text{ m/s}$
Plug in the values:
$F_d = 6 \pi (0.001 \text{ Pa} \cdot \text{s}) (0.0005 \text{ m}) (0.1 \text{ m/s})$
$F_d โ 9.42 \times 10^{-7} \text{ N}$
The drag force on the particle is approximately $9.42 \times 10^{-7}$ N.
Problem 2: Finding Viscosity
A small ball with a radius of 2 mm falls through a liquid at a constant velocity of 0.05 m/s. The drag force is measured to be $3 \times 10^{-6}$ N. Determine the viscosity of the liquid.
Solution:
Using Stokes' Law:
$F_d = 6 \pi \eta r v$
Rearrange to solve for $\eta$:
$\eta = \frac{F_d}{6 \pi r v}$
Given:
- ๐ช $F_d = 3 \times 10^{-6} \text{ N}$
- ๐ $r = 2 \text{ mm} = 0.002 \text{ m}$
- ๐ $v = 0.05 \text{ m/s}$
Plug in the values:
$\eta = \frac{3 \times 10^{-6} \text{ N}}{6 \pi (0.002 \text{ m}) (0.05 \text{ m/s})}$
$\eta โ 0.00159 \text{ Pa} \cdot \text{s}$
The viscosity of the liquid is approximately 0.00159 Paยทs.
Problem 3: Calculating Terminal Velocity
A spherical particle with a radius of 1 mm and a density of 1500 kg/mยณ is falling through oil with a density of 800 kg/mยณ and a viscosity of 0.1 Paยทs. Calculate the terminal velocity.
Solution:
At terminal velocity, the drag force equals the effective weight of the particle:
$F_d = W_{\text{effective}} = V (
ho_{\text{particle}} -
ho_{\text{fluid}}) g$
Where:
- ๐ฆ $V = \frac{4}{3} \pi r^3$ (volume of the sphere)
- ๐ $\rho_{\text{particle}} = 1500 \text{ kg/m}^3$
- ๐ข๏ธ $\rho_{\text{fluid}} = 800 \text{ kg/m}^3$
- ๐ $g = 9.81 \text{ m/s}^2$
First, calculate the volume:
$V = \frac{4}{3} \pi (0.001 \text{ m})^3 โ 4.19 \times 10^{-9} \text{ m}^3$
Now, calculate the effective weight:
$W_{\text{effective}} = (4.19 \times 10^{-9} \text{ m}^3) (1500 \text{ kg/m}^3 - 800 \text{ kg/m}^3) (9.81 \text{ m/s}^2) โ 2.87 \times 10^{-5} \text{ N}$
Using Stokes' Law, $F_d = 6 \pi \eta r v$:
$v = \frac{F_d}{6 \pi \eta r} = \frac{2.87 \times 10^{-5} \text{ N}}{6 \pi (0.1 \text{ Pa} \cdot \text{s}) (0.001 \text{ m})}$
$v โ 0.0152 \text{ m/s}$
The terminal velocity is approximately 0.0152 m/s.
๐ฏ Conclusion
Stokes' Law provides a fundamental understanding of drag forces on small objects in viscous fluids. It's essential in various scientific and engineering applications. By understanding its principles and applications, you can solve complex problems related to fluid dynamics. Keep practicing and exploring!