Let's solve the problem **step by step** as requested. ### Given Data - **Sample mass:** 100 g = 0.1 kg - **Density (\(\rho\)):** 7200 kg/m³ - **Sphericity (\(\phi_s\)):** 0.825 #### Screen Analysis Table | Mesh | Avg. Particle Diam. (mm) | Mass Retained (g) | |------|-------------------------|-------------------| | 8 | 2.845 | 0.00 | | 10 | 2.007 | 60.50 | | 14 | 1.409 | 27.25 | | 20 | 1.001 | 12.00 | | Pan | - | 0.25 | ### 1. For material between 10 mesh to 20 mesh: That includes the fractions: - 10 mesh (2.007 mm): 60.50 g - 14 mesh (1.409 mm): 27.25 g - 20 mesh (1.001 mm): 12.00 g #### Sum of mass in this range: \[ \text{Total mass (10-20 mesh)} = 60.50 + 27.25 + 12.00 = 99.75\,g = 0.09975\,kg \] --- ## (i) **Calculate \(A_\omega\) in m²/kg** ### What is \(A_\omega\)? \(A_\omega\) is the **specific surface area** (surface area per unit mass). For a particle: \[ A_\omega = \frac{\text{Total surface area of all particles}}{\text{Total mass}} \] For each size fraction: - **Particle volume:** \(V = \frac{\pi}{6} d^3\) (for a sphere) - **Particle surface area:** \(A = \pi d^2\) - For non-spherical particles, use sphericity (\(\phi_s\)): - Real surface area \(= \phi_s \times\) surface area of sphere of same volume. But for screen analysis, the **specific surface area** is: \[ A_\omega = \frac{6}{\rho \cdot d_s} \] where \(d_s\) is the **volume-surface mean diameter**. But often, for a size range, we take the weighted average: \[ A_\omega = \frac{\sum (A_i)}{\sum (m_i)} \] But since we are given the particle size for each fraction, let's compute for the mixture between 10-20 mesh. ### Step 1: For each size, calculate number of particles For each fraction: \[ \text{Particle mass} = V \cdot \rho = \frac{\pi}{6} d_p^3 \cdot \rho \] \[ \text{Number of particles} = \frac{\text{Mass retained}}{\text{mass per particle}} \] Let's calculate for each mesh: #### **A. 10 mesh (2.007 mm, 60.50 g)** - \(d_p = 2.007\,\text{mm} = 2.007 \times 10^{-3}\,\text{m}\) - \(m = 60.50\,\text{g} = 0.0605\,\text{kg}\) \[ \text{Particle volume} = \frac{\pi}{6} d_p^3 = \frac{3.1416}{6} \times (2.007 \times 10^{-3})^3 = 0.5236 \times (8.09 \times 10^{-9}) \approx 4.236 \times 10^{-9}\,\text{m}^3 \] \[ \text{Particle mass} = 4.236 \times 10^{-9} \times 7200 = 3.050 \times 10^{-5}\,\text{kg} \] \[ \text{Number of particles} = \frac{0.0605}{3.050 \times 10^{-5}} = 1.98 \times 10^{3} \] #### **B. 14 mesh (1.409 mm, 27.25 g)** - \(d_p = 1.409 \times 10^{-3}\,\text{m}\) - \(m = 0.02725\,\text{kg}\) \[ \text{Particle volume} = 0.5236 \times (1.409 \times 10^{-3})^3 = 0.5236 \times 2.799 \times 10^{-9} \approx 1.465 \times 10^{-9}\,\text{m}^3 \] \[ \text{Particle mass} = 1.465 \times 10^{-9} \times 7200 = 1.055 \times 10^{-5}\,\text{kg} \] \[ \text{Number of particles} = \frac{0.02725}{1.055 \times 10^{-5}} = 2.584 \times 10^{3} \] #### **C. 20 mesh (1.001 mm, 12.00 g)** - \(d_p = 1.001 \times 10^{-3}\,\text{m}\) - \(m = 0.012\,\text{kg}\) \[ \text{Particle volume} = 0.5236 \times (1.001 \times 10^{-3})^3 = 0.5236 \times 1.003 \times 10^{-9} \approx 5.251 \times 10^{-10}\,\text{m}^3 \] \[ \text{Particle mass} = 5.251 \times 10^{-10} \times 7200 = 3.781 \times 10^{-6}\,\text{kg} \] \[ \text{Number of particles} = \frac{0.012}{3.781 \times 10^{-6}} = 3.176 \times 10^{3} \] --- ### Step 2: Surface area for each fraction \[ \text{Surface area per particle} = \pi d_p^2 \] \[ \text{Real surface area per particle} = \phi_s \pi d_p^2 \] \[ \text{Total surface area (fraction)} = \text{Number of particles} \times \phi_s \pi d_p^2 \] Let's calculate: #### **10 mesh** \[ \pi d_p^2 = 3.1416 \times (2.007 \times 10^{-3})^2 = 3.1416 \times 4.028 \times 10^{-6} = 1.266 \times 10^{-5} \] \[ \text{Surface area per particle} = 0.825 \times 1.266 \times 10^{-5} = 1.044 \times 10^{-5} \] \[ \text{Total surface area} = 1.98 \times 10^3 \times 1.044 \times 10^{-5} = 0.02067\,\text{m}^2 \] #### **14 mesh** \[ \pi d_p^2 = 3.1416 \times (1.409 \times 10^{-3})^2 = 3.1416 \times 1.987 \times 10^{-6} = 6.247 \times 10^{-6} \] \[ \text{Surface area per particle} = 0.825 \times 6.247 \times 10^{-6} = 5.152 \times 10^{-6} \] \[ \text{Total surface area} = 2.584 \times 10^3 \times 5.152 \times 10^{-6} = 0.0133\,\text{m}^2 \] #### **20 mesh** \[ \pi d_p^2 = 3.1416 \times (1.001 \times 10^{-3})^2 = 3.1416 \times 1.002 \times 10^{-6} = 3.147 \times 10^{-6} \] \[ \text{Surface area per particle} = 0.825 \times 3.147 \times 10^{-6} = 2.595 \times 10^{-6} \] \[ \text{Total surface area} = 3.176 \times 10^3 \times 2.595 \times 10^{-6} = 0.00824\,\text{m}^2 \] --- ### Step 3: Add up for total (10-20 mesh): \[ \text{Total surface area} = 0.02067 + 0.0133 + 0.00824 = 0.0422\,\text{m}^2 \] \[ A_\omega = \frac{0.0422}{0.09975} = 0.423\,\text{m}^2/\text{kg} \] --- ## (ii) **Volume-surface mean diameter (\(d_{vs}\), in SI units)** \[ d_{vs} = \frac{\sum n_i d_{p,i}^3}{\sum n_i d_{p,i}^2} \] Or, using mass and density, \[ d_{vs} = \frac{\sum m_i d_{p,i}}{\sum m_i} \] But for more accuracy, use number of particles: \[ d_{vs} = \frac{\sum n_i d_{p,i}^3}{\sum n_i d_{p,i}^2} \] From above, \(n_i\): - 10 mesh: \(1.98 \times 10^3\), \(d_{p,1} = 2.007 \times 10^{-3}\) - 14 mesh: \(2.584 \times 10^3\), \(d_{p,2} = 1.409 \times 10^{-3}\) - 20 mesh: \(3.176 \times 10^3\), \(d_{p,3} = 1.001 \times 10^{-3}\) Calculate numerator: \[ \sum n_i d_{p,i}^3 = (1.98 \times 10^3) \times (2.007 \times 10^{-3})^3 + (2.584 \times 10^3) \times (1.409 \times 10^{-3})^3 + (3.176 \times 10^3) \times (1.001 \times 10^{-3})^3 \] Calculate each term: - 10 mesh: \(1.98 \times 10^3 \times 8.09 \times 10^{-9} = 1.602 \times 10^{-5}\) - 14 mesh: \(2.584 \times 10^3 \times 2.799 \times 10^{-9} = 7.233 \times 10^{-6}\) - 20 mesh: \(3.176 \times 10^3 \times 1.003 \times 10^{-9} = 3.184 \times 10^{-6}\) Sum: \[ \text{Numerator} = 1.602 \times 10^{-5} + 7.233 \times 10^{-6} + 3.184 \times 10^{-6} = 2.644 \times 10^{-5} \] Denominator: \[ \sum n_i d_{p,i}^2 = (1.98 \times 10^3) \times (2.007 \times 10^{-3})^2 + (2.584 \times 10^3) \times (1.409 \times 10^{-3})^2 + (3.176 \times 10^3) \times (1.001 \times 10^{-3})^2 \] - 10 mesh: \(1.98 \times 10^3 \times 4.028 \times 10^{-6} = 7.975 \times 10^{-3}\) - 14 mesh: \(2.584 \times 10^3 \times 1.987 \times 10^{-6} = 5.134 \times 10^{-3}\) - 20 mesh: \(3.176 \times 10^3 \times 1.002 \times 10^{-6} = 3.181 \times 10^{-3}\) Sum: \[ \text{Denominator} = 7.975 \times 10^{-3} + 5.134 \times 10^{-3} + 3.181 \times 10^{-3} = 1.629 \times 10^{-2} \] \[ d_{vs} = \frac{2.644 \times 10^{-5}}{1.629 \times 10^{-2}} = 1.623 \times 10^{-3}\,\text{m} = 1.62\,\text{mm} \] --- ## (iii) **Total number of particles (including pan)** Include all fractions except for 8 mesh (mass = 0): - 10 mesh: \(1.98 \times 10^{3}\) - 14 mesh: \(2.584 \times 10^{3}\) - 20 mesh: \(3.176 \times 10^{3}\) - Pan: For pan: - Mass = 0.25 g = 0.00025 kg - Assume d is much

Let's carefully work through each part of the problem step-by-step. --- ### **Given Data** | Parameter | Value | |--------------|---------| | Sample mass \(m_{total}\) | 100 g = 0.1 kg | | Density \(\rho\) | 7200 kg/m³ | | Sphericity \(\phi_s\) | 0.825 | ### **Screen Analysis Table** | Mesh | Avg. Particle Diameter \(d_i\) (mm) | Mass Retained \(m_i\) (g) | |---------|------------------------------|-------------------------| | 8 | 2.845 | 0.00 | | 10 | 2.007 | 60.50 | | 14 | 1.409 | 27.25 | | 20 | 1.001 | 12.00 | | Pan | - | 0.25 | --- ## Step 1: Identify the material between **10 mesh** and **20 mesh** Masses in this range: \[ m_{10-20} = 60.50 + 27.25 + 12.00 = 99.75\, \text{g} = 0.09975\, \text{kg} \] --- ## (i) **Calculate \(A_\omega\) in m²/kg (Specific Surface Area)** ### Concept: \[ A_\omega = \frac{\text{Total surface area of particles}}{\text{Total mass}} \] Since particles are approximately spherical, the surface area per particle: \[ A_{particle} = \pi d^2 \] and the **real surface area** considering sphericity: \[ A_{real} = \phi_s \times \pi d^2 \] The total surface area for a size fraction: \[ A_{total} = N \times A_{real} \] where \(N\) is the total number of particles in that fraction. ### Step 1: Calculate the number of particles \(N_i\) in each fraction: \[ N_i = \frac{m_i}{m_{particle,i}} \] where the mass of each particle: \[ m_{particle,i} = \frac{\pi}{6} d_i^3 \times \rho \] **Convert diameters to meters:** | \(d_i\) (mm) | \(d_i\) (m) | |--------------|--------------| | 2.007 | \(2.007 \times 10^{-3}\) | | 1.409 | \(1.409 \times 10^{-3}\) | | 1.001 | \(1.001 \times 10^{-3}\) | ### Particle masses: - For 10 mesh: \[ m_{particle,10} = \frac{\pi}{6} (2.007 \times 10^{-3})^3 \times 7200 \] Calculate: \[ (2.007 \times 10^{-3})^3 = 8.09 \times 10^{-9} \] \[ m_{particle,10} = \frac{\pi}{6} \times 8.09 \times 10^{-9} \times 7200 \approx 4.236 \times 10^{-5}\, \text{kg} \] Similarly for 14 and 20 mesh: - 14 mesh: \[ (1.409 \times 10^{-3})^3 = 2.80 \times 10^{-9} \] \[ m_{particle,14} \approx 1.465 \times 10^{-5}\, \text{kg} \] - 20 mesh: \[ (1.001 \times 10^{-3})^3 \approx 1.00 \times 10^{-9} \] \[ m_{particle,20} \approx 5.251 \times 10^{-6}\, \text{kg} \] ### Number of particles: | Fraction | \(N_i\) | |------------|--------------| | 10 mesh | \(\frac{0.0605}{4.236 \times 10^{-5}} \approx 1,429\) | | 14 mesh | \(\frac{0.2725}{1.465 \times 10^{-5}} \approx 18,607\) | | 20 mesh | \(\frac{0.1200}{5.251 \times 10^{-6}} \approx 22,842\) | ### Step 2: Calculate total surface area for each: \[ A_{particle,i} = \pi d_i^2 \] - 10 mesh: \[ A_{particle,10} = \pi \times (2.007 \times 10^{-3})^2 \approx 1.266 \times 10^{-5}\, \text{m}^2 \] - 14 mesh: \[ A_{particle,14} = 3.1416 \times (1.409 \times 10^{-3})^2 \approx 6.247 \times 10^{-6}\, \text{m}^2 \] - 20 mesh: \[ A_{particle,20} = 3.1416 \times (1.001 \times 10^{-3})^2 \approx 3.147 \times 10^{-6}\, \text{m}^2 \] ### Step 3: Total surface area for each: Apply sphericity: \[ A_{real,i} = \phi_s \times A_{particle,i} \] Total surface area: | Fraction | Total surface area \(A_{total,i}\) (m²) | |------------|------------------------------------------| | 10 mesh | \(1,429 \times 0.825 \times 1.266 \times 10^{-5} \approx 0.0149 \, \text{m}^2 \) | | 14 mesh | \(18,607 \times 0.825 \times 6.247 \times 10^{-6} \approx 0.096 \, \text{m}^2 \) | | 20 mesh | \(22,842 \times 0.825 \times 3.147 \times 10^{-6} \approx 0.059 \, \text{m}^2 \) | Sum total: \[ A_{total} = 0.0149 + 0.096 + 0.059 \approx 0.17\, \text{m}^2 \] --- ### **Final \(A_\omega\):** \[ A_\omega = \frac{A_{total}}{m_{total}} = \frac{0.17}{0.09975} \approx \boxed{1.70\, \text{m}^2/\text{kg}} \] --- ## (ii) **Volume-surface mean diameter (\(d_{vs}\))** \[ d_{vs} = \frac{\sum n_i d_i^3}{\sum n_i d_i^2} \] Using previously calculated: \[ \sum n_i d_i^3 = 2.644 \times 10^{-5} \] \[ \sum n_i d_i^2 = 1.629 \times 10^{-2} \] Calculate: \[ d_{vs} = \frac{2.644 \times 10^{-5}}{1.629 \times 10^{-2}} = 1.623 \times 10^{-3}\, \text{m} = 1.62\, \text{mm} \] --- ## (iii) **Total number of particles including pan** Total particles across all fractions: \[ N_{total} = N_{10} + N_{14} + N_{20} + N_{pan} \] - \(N_{10} \approx 1,429\) - \(N_{14} \approx 18,607\) - \(N_{20} \approx 22,842\) - Pan: 0.25 g = \(2.5 \times 10^{-4}\) kg Number of particles in pan: \[ m_{pan} = 2.5 \times 10^{-4}\, \text{kg} \] \[ N_{pan} = \frac{2.5 \times 10^{-4}}{m_{particle,pan}} \] Estimate \(m_{particle,pan}\): - Assume average particle diameter in pan is very small; for simplicity, neglect the pan particles or assume they are negligible compared to the other fractions. **Alternatively**, since the total mass is 0.1 kg, and the sum of retained masses is 0.09975 kg, the pan's contribution is small and the number of particles is negligible in comparison. **Final total particles:** \[ N_{total} \approx 1,429 + 18,607 + 22,842 \approx \boxed{42,878} \] --- ### **Final answers:** | Part | Result | |-------|---------| | (i) \(A_\omega\) | **1.70 m²/kg** | | (ii) Volume-surface mean diameter \(d_{vs}\) | **1.62 mm** | | (iii) Total number of particles | **Approximately 42,878** | --- **Note:** For precise calculations, some assumptions were made regarding the pan particles. The main focus was on the fractions between 10 and 20 mesh. --- **If you need further calculations or clarifications, feel free to ask!**