Let's break down the problem step by step using your data and the required calculations: ### **Given Data** | Mesh Size | Avg. Particle Diameter (mm) | Mass Retained (g) | |-----------|-----------------------------|-------------------| | 8 | 2.845 | 0.00 | | 10 | 2.007 | 60.5 | | 14 | 1.409 | 27.25 | | 20 | 1.001 | 12.00 | | Pan | - | 0.25 | - **Total mass:** \( 100 \) gm - **Density (\(\rho\))**: \( 7200 \) kg/m³ - **Sphericity (\(\phi_s\))**: 0.825 Let's calculate for the material **between 10 mesh and 20 mesh** (i.e., sum of 10, 14, and 20 mesh fractions). --- ## **Step 1: Calculate total mass between 10 mesh and 20 mesh** \[ \text{Total mass (10-20 mesh)} = 60.5 + 27.25 + 12.00 = 99.75\, \text{gm} \] We'll use 99.75 gm for all the further calculations. --- ## **Step 2: Calculate \(A_\omega\) (Surface area per unit mass, m²/kg)** ### **a) Calculate surface area for each size fraction** For each size, the surface area of a single particle: \[ \text{Surface area of sphere} = \pi d^2 \] But with sphericity, the actual surface area: \[ \text{Surface area of one particle} = \frac{\pi d^2}{\phi_s} \] However, the formula for the total surface area of all particles in a fraction is (using SI units): \[ \text{Number of particles in fraction } n_i = \frac{\text{mass}_i}{\text{volume of 1 particle} \times \rho} \] \[ \text{Volume of one particle} = \frac{\pi}{6} d^3 \] So, \[ n_i = \frac{m_i}{\frac{\pi}{6} d^3 \rho} \] Total surface area for each fraction: \[ A_i = n_i \cdot \frac{\pi d^2}{\phi_s} \] \[ A_i = \frac{m_i}{\frac{\pi}{6} d^3 \rho} \cdot \frac{\pi d^2}{\phi_s} \] \[ A_i = \frac{m_i \cdot 6}{d^3 \rho} \cdot \frac{d^2}{\phi_s} \] \[ A_i = \frac{6 m_i}{d \rho \phi_s} \] **Sum for all three mesh fractions (10, 14, 20):** \[ A_\text{total} = \sum \frac{6 m_i}{d_i \rho \phi_s} \] ### **b) Plug in values for each fraction** Convert all diameters to **meters** and masses to **kg**. | Mesh | d (mm) | d (m) | m (gm) | m (kg) | |------|--------|-----------|--------|-----------| | 10 | 2.007 | 0.002007 | 60.5 | 0.0605 | | 14 | 1.409 | 0.001409 | 27.25 | 0.02725 | | 20 | 1.001 | 0.001001 | 12.00 | 0.01200 | Given: - \(\rho = 7200\, \text{kg/m}^3\) - \(\phi_s = 0.825\) Calculate each term: ### For Mesh 10: \[ A_{10} = \frac{6 \times 0.0605}{0.002007 \times 7200 \times 0.825} \] \[ = \frac{0.363}{11.93034} \] \[ = 0.03044\, \text{m}^2 \] ### For Mesh 14: \[ A_{14} = \frac{6 \times 0.02725}{0.001409 \times 7200 \times 0.825} \] \[ = \frac{0.1635}{8.38188} \] \[ = 0.01951\, \text{m}^2 \] ### For Mesh 20: \[ A_{20} = \frac{6 \times 0.01200}{0.001001 \times 7200 \times 0.825} \] \[ = \frac{0.072}{5.955} \] \[ = 0.01210\, \text{m}^2 \] ### **c) Sum total surface area** \[ A_\text{total} = 0.03044 + 0.01951 + 0.01210 = 0.06205\, \text{m}^2 \] ### **d) Surface area per unit mass (\(A_\omega\))** \[ A_\omega = \frac{A_\text{total}}{\text{Total mass}} \] Remember total mass in **kg**: \(0.0605 + 0.02725 + 0.01200 = 0.09975\, \text{kg}\) \[ A_\omega = \frac{0.06205}{0.09975} = 0.622\, \text{m}^2/\text{kg} \] --- ## **Step 3: Volume-surface mean diameter (\(\mathbf{d_{vs}}\))** \[ d_{vs} = \frac{\text{Total volume}}{\text{Total surface area}} \times 6 \] But more generally, \[ d_{vs} = \frac{\sum n_i d_i^3}{\sum n_i d_i^2} \] But with mass fractions, another formula is: \[ d_{vs} = \frac{\sum m_i d_i}{\sum m_i} \] But for **volume-surface mean diameter** (Sauter mean diameter): \[ d_{32} = \frac{\sum n_i d_i^3}{\sum n_i d_i^2} \] But since we can compute everything in terms of mass (because all particles are of same density and sphericity), we can use: \[ d_{32} = \frac{\sum m_i d_i}{\sum m_i \frac{d_i^2}{d_i}} = \frac{\sum m_i d_i}{\sum m_i} \] But this is the arithmetic mean based on mass; for the Sauter mean diameter for batch of different sizes, a more accurate formula is: \[ d_{32} = \frac{\sum m_i d_i}{\sum m_i \frac{d_i^2}{d_i}} \] But for spheres of same density, the Sauter mean diameter is: \[ d_{32} = \frac{6 V_\text{total}}{A_\text{total}} \] Where \(V_\text{total}\) is the total volume of particles. First, compute total volume: \[ \text{Volume of each fraction} = \frac{m_i}{\rho} \] \[ V_\text{total} = \frac{0.0605}{7200} + \frac{0.02725}{7200} + \frac{0.01200}{7200} \] \[ = \frac{0.09975}{7200} = 1.3854 \times 10^{-5} \, \text{m}^3 \] Now, \[ d_{32} = \frac{6 \times 1.3854 \times 10^{-5}}{0.06205} \] \[ = \frac{8.3124 \times 10^{-5}}{0.06205} \] \[ = 0.00134 \text{ m} = 1.34 \text{ mm} \] --- ## **Step 4: Total number of particles (including pan)** Use: \[ n_i = \frac{m_i}{\text{volume of one particle} \times \rho} \] Where volume of one particle: \[ V_p = \frac{\pi}{6} d^3 \] So, \[ n_i = \frac{m_i}{\frac{\pi}{6} d^3 \rho } = \frac{6 m_i}{\pi d^3 \rho} \] Calculate for each fraction (ignore 8 mesh as mass is zero, for pan assume d is very small, so number is negligible). | Mesh | m (kg) | d (m) | \(d^3\) | \(n_i\) | |------|--------|----------|---------------|------------------------------| | 10 | 0.0605 | 0.002007 | 8.094e-9 | \(n_{10} = \frac{6 \times 0.0605}{\pi \times 8.094e-9 \times 7200}\) | | 14 | 0.02725| 0.001409 | 2.799e-9 | \(n_{14} = \frac{6 \times 0.02725}{\pi \times 2.799e-9 \times 7200}\) | | 20 | 0.01200| 0.001001 | 1.003e-9 | \(n_{20} = \frac{6 \times 0.012}{\pi \times 1.003e-9 \times 7200}\) | **Calculate each:** ### Mesh 10 \[ n_{10} = \frac{0.363}{\pi \times 8.094 \times 10^{-9} \times 7200} \] \[ = \frac{0.363}{\pi \times 5.828 \times 10^{-5}} \] \[ = \frac{0.363}{1.831 \times 10^{-4}} \] \[ = 1983 \] ### Mesh 14 \[ n_{14} = \frac{0.1635}{\pi \times 2.799 \times 10^{-9} \times 7200} \] \[ = \frac{0.1635}{\pi \times 2.015 \times 10^{-5}} \] \[ = \frac{0.1635}{6.333 \times 10^{-5}} \] \[ = 2582 \] ### Mesh 20 \[ n_{20} = \frac{0.072}{\pi \times 1.003 \times 10^{-9} \times 7200} \] \[ = \frac{0.072}{\pi \times 7.221 \times 10^{-6}} \] \[ = \frac{0.072}{2.269 \times 10^{-5}} \] \[ = 3175 \] ### Pan: mass is only 0.25 gm = \(2.5 \times 10^{-4}\) kg, but diameter is not given, so number will be very large but negligible in total mass, so can be ignored. \[ \text{Total number of particles (including pan)} \approx 1983 + 2582 + 3175 = 7739 \] --- ## **Step 5: Volume Mean Diameter (\(\mathbf{d_{43}}\))** \[ d_{43} = \frac{\sum n_i d_i^4}{\sum n_i d_i^3} \] Calculate \(n_i d_i^4\) and \(n_i d_i^3\): | Mesh | \(n_i\) | \(d_i\) (m) | \(d_i^4\) | \(d_i^3\) | \(n_i d_i^4\) | \(n_i d_i^3\) | |------|--------|-------------|-----------|-----------|---------------|---------------| | 10 |1983 |0.002007 | 1.625e-11 | 8.094e-9 | 3.224e-8 | 1.606e-5 | | 14 |2582 |0.001409 | 3.938e-12 | 2.799e-9 | 1.016e-8 | 7.236e-6 | | 20 |3175 |0.001001 | 1.004e-12 | 1.003e-9 | 3.187e-9 | 3.183e-6 | Sum: \[ \sum n_i d_i^4 = 3.224e-8 + 1.016e-8 + 3.187e-9 = 4.559e-8 \] \[ \sum n_i d_i^3 = 1.606e-5 + 7.236e-6 + 3.183e-6 = 2.648e-5 \] So, \[ d_{43} = \frac{4.559 \times 10^{-8}}{2.648 \times 10