Let's solve the problem step by step using the data and requirements you've provided. --- ### **Given Data** - **Sample mass:** \( 100 \, \text{gm} \) (Assuming the "1 0 0" in your statement is 100 gm) - **True density (\( \rho \))**: \( 7200 \, \text{kg/m}^3 \) - **Sphericity (\( \phi_1 \)):** \( 0.825 \) - **Screen Analysis Table:** | Mesh Size | Avg. Particle Diameter (mm) | Mass Retained (gm) | |-----------|-----------------------------|---------------------| | 10 | 2.845 | 0.00 | | 14 | 2.007 | 60.5 | | 20 | 1.409 | 27.25 | | Pan | 1.001 | 12.00 | | (Total) | | 99.75 (∼100) | --- ## **Step 1: For material between 10 mesh and 20 mesh** This means **material retained on 14 mesh and 20 mesh**. So, **mass between 10 and 20 mesh** = **60.5 gm (14 mesh) + 27.25 gm (20 mesh) = 87.75 gm** --- ## **Step 2: Calculate Specific Surface Area (A) in \( m^2/kg \) for this fraction** **Specific surface area (A):** \[ A = \frac{\text{Surface area of all particles}}{\text{Mass of all particles}} \] For a single particle: \[ \text{Surface area} = \phi_1 \times (\text{surface area of sphere}) = \phi_1 \times \pi d^2 \] \[ \text{Volume} = \frac{\pi}{6} d^3 \] \[ \text{Mass} = \text{Volume} \times \rho \] But for average specific surface area, we use the **surface-volume mean diameter** (\( d_{sv} \)): \[ A = \frac{6}{\phi_1 \cdot \rho \cdot d_{sv}} \] But since the size range is not narrow, we use the following formulation for the **fraction** (for each size range): \[ A = \frac{\sum \left( \frac{\text{Mass fraction}}{\text{density}} \times \frac{6}{\phi_1 \cdot d} \right)}{\sum \text{mass fraction}/\rho} \] But since all particles have the same density and sphericity, for a mixture: \[ A = \frac{\sum \left( \frac{m_i}{d_i} \right)}{\rho \cdot \phi_1 \cdot \sum m_i} \] Where - \( m_i \) = mass of fraction \( i \) - \( d_i \) = average diameter of fraction \( i \) **For 10 to 20 mesh:** - 14 mesh: \( d_{14} = 2.007 \) mm, \( m_{14} = 60.5 \) gm - 20 mesh: \( d_{20} = 1.409 \) mm, \( m_{20} = 27.25 \) gm \[ \sum m_i = 60.5 + 27.25 = 87.75 \text{ gm} \] \[ \sum \left( \frac{m_i}{d_i} \right) = \frac{60.5}{2.007} + \frac{27.25}{1.409} \] Calculate: - \( \frac{60.5}{2.007} \approx 30.15 \) - \( \frac{27.25}{1.409} \approx 19.35 \) - Total: \( 30.15 + 19.35 = 49.5 \) So, \[ A = \frac{49.5}{7200 \times 0.825 \times 87.75 \times 10^{-3}} \] Convert gm to kg for mass: \( 87.75 \) gm = \( 0.08775 \) kg - \( \rho \) in kg/m³ - \( m \) in kg So, \[ A = \frac{49.5}{7200 \times 0.825 \times 0.08775} \] Calculate denominator: - \( 7200 \times 0.825 = 5940 \) - \( 5940 \times 0.08775 \approx 521.115 \) So, \[ A = \frac{49.5}{521.115} = 0.095 \ \text{m}^2/\text{kg} \] --- ## **Step 3: Volume-surface mean diameter (in SI)** **Definition:** \[ d_{sv} = \frac{\sum (m_i)}{\sum \left( \frac{m_i}{d_i} \right)} \] \[ d_{sv} = \frac{87.75}{49.5} \] But, units must be consistent (diameters in mm, but we want SI, so convert to meters): \[ d_{sv} = \frac{87.75}{49.5} \ \text{mm} = 1.774 \ \text{mm} = 1.774 \times 10^{-3} \ \text{m} \] --- ## **Step 4: Total number of particles in pan** - Pan: \( d_{pan} = 1.001 \) mm - \( m_{pan} = 12.00 \) gm = 0.012 kg **For one particle:** \[ \text{Volume} = \frac{\pi}{6} d^3 \ (\text{for a sphere}) \] \[ \text{Mass} = \text{Volume} \times \rho \] So, number of particles: \[ N = \frac{\text{Total mass}}{\text{Mass per particle}} \] Calculate for pan: - \( d = 1.001 \) mm = \( 1.001 \times 10^{-3} \) m \[ \text{Volume per particle} = \frac{\pi}{6} d^3 = \frac{\pi}{6} \times (1.001 \times 10^{-3})^3 = \frac{3.1416}{6} \times (1.003 \times 10^{-9}) \] \[ = 0.5236 \times 1.003 \times 10^{-9} = 0.524 \times 10^{-9} \text{ m}^3 \] \[ \text{Mass per particle} = 0.524 \times 10^{-9} \times 7200 = 3.77 \times 10^{-6} \text{ kg} \] \[ N = \frac{0.012}{3.77 \times 10^{-6}} \approx 3183 \] --- ## **Step 5: Volume mean diameter (\( d_{vm} \), in SI unit)** **Definition:** \[ d_{vm} = \frac{\sum (m_i d_i)}{\sum m_i} \] For the 10 to 20 mesh fraction: - \( \sum (m_i d_i) = (60.5 \times 2.007) + (27.25 \times 1.409) \) - \( = 121.2 + 38.42 = 159.62 \) - \( \sum m_i = 87.75 \) So, \[ d_{vm} = \frac{159.62}{87.75} = 1.819 \text{ mm} = 1.819 \times 10^{-3} \text{ m} \] --- ## **Summary Table** | Quantity | Result | |---------------------------|------------------------------| | (i) Specific surface area | \( 0.095 \ \text{m}^2/\text{kg} \) | | (ii) Volume-surface mean diameter (\( d_{sv} \)) | \( 1.774 \times 10^{-3} \ \text{m} \) | | (iii) Total particles in pan | \( 3180 \) | | (iv) Volume mean diameter (\( d_{vm} \)) | \( 1.819 \times 10^{-3} \ \text{m} \) | --- ## **Final Answers** **(i) Specific surface area, \( A \):** \[ A = 0.095 \ \text{m}^2/\text{kg} \] **(ii) Volume-surface mean diameter, \( d_{sv} \):** \[ d_{sv} = 1.77 \times 10^{-3} \ \text{m} \] **(iii) Total number of particles in pan:** \[ N = 3180 \] **(iv) Volume mean diameter, \( d_{vm} \):** \[ d_{vm} = 1.82 \times 10^{-3} \ \text{m} \] --- ### **If you need detailed calculation steps for any item, let me know!**