Terminal Settling Velocity of a Glass Cube

 

Illustrative Example 4.4: Terminal Settling Velocity of a Glass Cube

Problem Statement:
Calculate the terminal settling velocity of a glass cube with an edge dimension of 0.1 mm in a fluid with the following properties:

• Density of fluid ($\rho_f$) = 982 kg/m³
• Viscosity of fluid ($\mu$) = 0.0013 kg/m·s
• Density of glass ($\rho_s$) = 2820 kg/m³

Additionally, calculate:

1. The equivalent volume diameter ($d_v$) of the cube.
2. The sphericity factor ($\phi$) of the cube.

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Solution

Step 1: Calculate the Equivalent Volume Diameter ($d_v$)

The equivalent volume diameter ($d_v$) is the diameter of a sphere with the same volume as the cube.

1. Volume of the cube:

   $$V_{\text{cube}} = a^3 = (0.0001)^3 = 1 \times 10^{-12} \, \text{m}^3$$

2. Volume of a sphere:

   $$V_{\text{sphere}} = \frac{\pi}{6} d_v^3$$

3. Equating the two volumes:

   $$1 \times 10^{-12} = \frac{\pi}{6} d_v^3$$ $$d_v^3 = \frac{6 \cdot 1 \times 10^{-12}}{\pi} = 1.91 \times 10^{-12} \, \text{m}^3$$ $$d_v = \sqrt[3]{1.91 \times 10^{-12}} = 0.000125 \, \text{m} \, \text{(or 0.125 mm)}$$

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Step 2: Calculate the Sphericity Factor ($\phi$)

Sphericity ($\phi$) is the ratio of the surface area of a sphere with the same volume to the surface area of the cube.

1. Surface area of the cube:

   $$A_{\text{cube}} = 6 a^2 = 6 (0.0001)^2 = 6 \times 10^{-8} \, \text{m}^2$$

2. Surface area of the sphere:

   $$A_{\text{sphere}} = \pi d_v^2 = \pi (0.000125)^2 = 4.91 \times 10^{-8} \, \text{m}^2$$

3. Sphericity:

   $$\phi = \frac{A_{\text{sphere}}}{A_{\text{cube}}} = \frac{4.91 \times 10^{-8}}{6 \times 10^{-8}} = 0.818$$

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Step 3: Calculate the Terminal Settling Velocity ($v_t$)

Stokes' law for terminal settling velocity is:

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

Where $r_v = \frac{d_v}{2}$:

$$r_v = \frac{0.000125}{2} = 0.0000625 \, \text{m}$$

Substitute the known values:

$$v_t = \frac{2 (2820 - 982) (9.81) (0.0000625)^2}{9 (0.0013)}$$

1. Density difference:

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

2. Square the radius:

   $$r_v^2 = (0.0000625)^2 = 3.91 \times 10^{-9} \, \text{m}^2$$

3. Numerator:

   $$2 \cdot 1838 \cdot 9.81 \cdot 3.91 \times 10^{-9} = 1.41 \times 10^{-4}$$

4. Denominator:

   $$9 \cdot 0.0013 = 0.0117$$

5. Terminal velocity:

   $$v_t = \frac{1.41 \times 10^{-4}}{0.0117} = 0.0121 \, \text{m/s}$$

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Results

1. Equivalent Volume Diameter ($d_v$): $0.000125 \, \text{m} \, \text{(or 0.125 mm)}$
2. Sphericity Factor ($\phi$): $0.818$
3. Terminal Settling Velocity ($v_t$): $0.0121\text{m/s}\text{(or 12.1 mm/s)}$