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Viscosity and Stokes' Law



Viscosity

In the previous section on fluid dynamics, we discussed ideal fluids which are non-viscous. However, real fluids exhibit viscosity, which is a measure of the internal friction within the fluid. Viscosity is the property of a fluid that opposes the relative motion between different layers of the fluid. It can be thought of as the "thickness" or "stickiness" of the fluid; highly viscous fluids (like honey or tar) flow slowly, while low viscosity fluids (like water or air) flow easily.


Viscous Force

When a fluid flows, different layers of the fluid move at different velocities. Viscosity gives rise to internal frictional forces (viscous forces) between these layers that oppose this relative motion. These forces are analogous to the friction between solid surfaces in relative motion.

Consider a layer of fluid flowing over another layer. The faster-moving layer exerts a forward tangential force on the slower-moving layer, while the slower-moving layer exerts a backward tangential force on the faster-moving layer. These forces are viscous forces.


Coefficient of Viscosity ($\eta$)

To quantify viscosity, we define the coefficient of viscosity (often denoted by $\eta$, sometimes $\mu$). Consider a fluid flowing between two parallel plates, where the bottom plate is stationary and the top plate moves with a velocity $v$. The fluid layers near the bottom plate are almost stationary, while the layers near the top plate move with the velocity of the top plate. This creates a velocity gradient perpendicular to the flow direction.

If the distance between the layers is $dz$ and the velocity difference between them is $dv$, the velocity gradient is $\frac{dv}{dz}$. Experiments show that the tangential viscous force $F_v$ between two layers of area $A$ is directly proportional to the area of the layers and the velocity gradient.

$ F_v \propto A \frac{dv}{dz} $

Introducing the constant of proportionality, the coefficient of viscosity $\eta$:

$ F_v = \eta A \frac{dv}{dz} $

This equation is known as Newton's law of viscosity, and fluids that obey this relationship are called Newtonian fluids (e.g., water, air). Some fluids (non-Newtonian fluids like ketchup, paint) have more complex relationships between shear stress and shear rate.

From the definition $ \eta = \frac{F_v/A}{dv/dz} = \frac{\text{Shear Stress}}{\text{Velocity Gradient}} $, the units of viscosity are Pascal-second (Pa·s) in SI units. $1 \text{ Pa} \cdot \text{s} = 1 \text{ N/m}^2 \cdot \text{s} = 1 \text{ kg/(m} \cdot \text{s})$.

A common CGS unit is the poise (P), named after Jean Léonard Marie Poiseuille. $1 \text{ poise} = 0.1 \text{ Pa} \cdot \text{s}$. Centipoise (cP) is often used, where $1 \text{ cP} = 0.01 \text{ poise} = 10^{-3} \text{ Pa} \cdot \text{s}$. The viscosity of water at 20°C is approximately 1 cP.


Dependence on Temperature

Viscosity is highly dependent on temperature.


Stokes’ Law ($ F_v = 6\pi\eta r v $)

When a small object moves through a viscous fluid, it experiences a viscous drag force that opposes its motion. For a small sphere moving slowly through a viscous fluid in laminar flow, the viscous drag force is given by Stokes' Law.

Stokes' Law states that the viscous drag force ($F_v$) on a small sphere of radius $r$ moving with a velocity $v$ through a fluid of viscosity $\eta$ is directly proportional to the radius of the sphere, its velocity, and the viscosity of the fluid.

Mathematically, the formula is:

$ F_v = 6\pi\eta r v $

where:

This formula is valid under certain conditions:


Terminal Velocity

When a solid object falls through a viscous fluid (like air or water), it is acted upon by three forces:

  1. Downward gravitational force (weight): $W = mg = \rho_{object} V_{object} g$.
  2. Upward buoyant force: $F_B = \rho_{fluid} V_{object} g$ (assuming full submersion/constant volume).
  3. Upward viscous drag force: $F_v = 6\pi\eta r v$ (for a sphere).

Initially, when the object is dropped from rest, its velocity is zero, so the viscous force is zero. As the object accelerates downwards, its speed increases, and the viscous force also increases. Eventually, the viscous force and buoyant force together become equal in magnitude to the gravitational force. At this point, the net force on the object is zero, and its acceleration becomes zero. The object then falls with a constant maximum velocity called the terminal velocity ($v_T$).

At terminal velocity, the forces are balanced:

$ W = F_B + F_v $

$ \rho_{object} V_{object} g = \rho_{fluid} V_{object} g + 6\pi\eta r v_T $

For a sphere of radius $r$, $V_{object} = \frac{4}{3}\pi r^3$.

$ \rho_{object} \left(\frac{4}{3}\pi r^3\right) g = \rho_{fluid} \left(\frac{4}{3}\pi r^3\right) g + 6\pi\eta r v_T $

$ \left(\rho_{object} - \rho_{fluid}\right) \frac{4}{3}\pi r^3 g = 6\pi\eta r v_T $

Solve for $v_T$:

$ v_T = \frac{(\rho_{object} - \rho_{fluid}) \frac{4}{3}\pi r^3 g}{6\pi\eta r} $

$ v_T = \frac{(\rho_{object} - \rho_{fluid}) \frac{4}{3} r^2 g}{6\eta} $

$ v_T = \frac{2 (\rho_{object} - \rho_{fluid}) r^2 g}{9 \eta} $

This formula gives the terminal velocity of a sphere falling through a viscous fluid. The terminal velocity depends on the densities of the object and fluid, the radius of the sphere (squared), gravity, and the viscosity of the fluid. Larger, denser objects fall faster in a viscous medium (have higher terminal velocity) than smaller, less dense objects. Higher fluid viscosity leads to lower terminal velocity.

Diagram showing forces on a sphere falling through a viscous fluid at terminal velocity.

(Image Placeholder: A sphere falling downwards in a fluid. Downward arrow for Weight W. Upward arrow for Buoyant Force FB. Upward arrow for Viscous Force Fv. At terminal velocity, W = FB + Fv.)


Applications of Viscosity and Stokes' Law

Example 1. A small steel ball of radius 2.0 mm and density $7.8 \times 10^3$ kg/m$^3$ falls through a liquid of viscosity $0.8 \, \text{Pa} \cdot \text{s}$ and density $1.2 \times 10^3$ kg/m$^3$. Calculate the terminal velocity of the ball. (Take $g = 9.8 \, \text{m/s}^2$).

Answer:

Radius of the steel ball, $r = 2.0$ mm $= 2.0 \times 10^{-3}$ m.

Density of steel, $\rho_{object} = 7.8 \times 10^3$ kg/m$^3$.

Viscosity of liquid, $\eta = 0.8$ Pa·s.

Density of liquid, $\rho_{fluid} = 1.2 \times 10^3$ kg/m$^3$.

$g = 9.8 \, \text{m/s}^2$.

Using the formula for terminal velocity of a sphere:

$ v_T = \frac{2 (\rho_{object} - \rho_{fluid}) r^2 g}{9 \eta} $

$ v_T = \frac{2 \times (7.8 \times 10^3 - 1.2 \times 10^3) \text{ kg/m}^3 \times (2.0 \times 10^{-3} \text{ m})^2 \times (9.8 \text{ m/s}^2)}{9 \times (0.8 \text{ Pa} \cdot \text{s})} $

$ v_T = \frac{2 \times (6.6 \times 10^3) \times (4.0 \times 10^{-6}) \times 9.8}{9 \times 0.8} $ m/s

$ v_T = \frac{13.2 \times 10^3 \times 4.0 \times 10^{-6} \times 9.8}{7.2} $ m/s

$ v_T = \frac{52.8 \times 9.8 \times 10^{-3}}{7.2} $ m/s

$ v_T \approx \frac{517.44 \times 10^{-3}}{7.2} $ m/s

$ v_T \approx 71.86 \times 10^{-3} $ m/s

$ v_T \approx 0.07186 $ m/s.

The terminal velocity of the steel ball in the liquid is approximately 0.072 m/s or 7.2 cm/s.