Calculation for Terminal Settling Velocity of a Glass Cube

 Illustrative example 4.4

Calculate the terminal settling velocity of a glass cube having edge dimension 0.1 mm in a fluid of density 982 kg/m3 and viscosity 0.0013 kg/ms. The density of the glass is 2820 kg3. Calculate the equivalent volume diameter and the sphericity factor.

Calculation for Terminal Settling Velocity of a Glass Cube

We are tasked with calculating the terminal settling velocity of a glass cube and determining its equivalent volume diameter and sphericity factor.

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Given Data

• Edge dimension of the cube ($a$) = 0.1 mm = $0.0001 \, \text{m}$
• Density of the cube ($\rho_s$) = $2820 \, \text{kg/m}^3$
• 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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1. Calculate the Equivalent Volume Diameter

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

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

The volume of a sphere is given by:

$$\text{Volume of the sphere} = \frac{\pi}{6} d_v^3$$

Equating the cube's volume to the sphere's volume:

$$1 \times 10^{-12} = \frac{\pi}{6} d_v^3$$

Solve for $d_v$:

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

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2. Calculate the Sphericity Factor

The sphericity factor ($\phi$) 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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3. Calculate the Terminal Settling Velocity Using Stokes' Law

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 values into the formula:

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

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: $v_t = 0.0121 \, \text{m/s} \, \text{(or 12.1 mm/s)}$

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