Calculation of Terminal Settling Velocity Using Stokes' Law

 Illustrative example 4.3

Use Stokes’ law to calculate the terminal settling velocity of a glass sphere of diameter 0.1 mm in a fluid having density 982 kg/m3 and viscosity 0.0013 kg/ms. Density of glass is 2820 kg/m3.

Calculation of Terminal Settling Velocity Using Stokes' Law

We are tasked with calculating the terminal settling velocity of a glass sphere using Stokes' law. Here's a step-by-step solution:

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Given Data

• Diameter of the sphere ($d$) = 0.1 mm = $0.0001 \, \text{m}$
• Radius of the sphere ($r$) = $\frac{d}{2} = 0.00005 \, \text{m}$
• Density of the sphere ($\rho_s$) = $2820 \, \text{kg/m}^3$
• Density of the fluid ($\rho_f$) = $982 \, \text{kg/m}^3$
• Dynamic viscosity of the fluid ($\mu$) = $0.0013 \, \text{kg/m·s}$
• Acceleration due to gravity ($g$) = $9.81 \, \text{m/s}^2$

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Stokes' Law for Terminal Settling Velocity

The terminal settling velocity ($v_t$) is given by:

$$v_t = \frac{2 (\rho_s - \rho_f) g r^2}{9 \mu}$$

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Substitute the Values

$$v_t = \frac{2 (2820 - 982) (9.81) (0.00005)^2}{9 (0.0013)}$$

1. Calculate the density difference:

   $$(\rho_s - \rho_f) = 2820 - 982 = 1838 \, \text{kg/m}^3$$

2. Square the radius:

   $$r^2 = (0.00005)^2 = 2.5 \times 10^{-9} \, \text{m}^2$$

3. Numerator:

   $$2 \cdot 1838 \cdot 9.81 \cdot 2.5 \times 10^{-9} = 9.005 \times 10^{-5}$$

4. Denominator:

   $$9 \cdot 0.0013 = 0.0117$$

5. Terminal Velocity:

   $$v_t = \frac{9.005 \times 10^{-5}}{0.0117} = 0.0077 \, \text{m/s}$$

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Final Answer

The terminal settling velocity of the glass sphere is approximately:

$$\boxed{0.0077 \, \text{m/s} \, \text{(or 7.7 mm/s)}}$$

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