Let's break down your problem and **design a cyclone separator** for **85% recovery** of solid particles from an air stream with a volumetric flow rate of **5000 m³/h** at **100°C**. The solid particle density is **2800 kg/m³**, and a particle size distribution is given. ### 1. **Convert Flow Rate to SI Units** \[ Q = 5000\, \text{m}^3/\text{h} = \frac{5000}{3600} = 1.39\, \text{m}^3/\text{s} \] --- ### 2. **Particle Size Distribution Table** | Size Range (μm) | Weight Fraction (%) | |-----------------|--------------------| | 60-50 | 6 | | 50-40 | 8 | | 40-30 | 12 | | 30-20 | 17 | | 20-10 | 25 | | 10-5 | 25 | | 5-2 | 5 | | 2-0 | 2 | --- ### 3. **Determine Cut Size for 85% Recovery** - **Cumulative Weight Fraction:** - 60-50 µm: 6% - 50-40 µm: 6 + 8 = 14% - 40-30 µm: 14 + 12 = 26% - 30-20 µm: 26 + 17 = 43% - 20-10 µm: 43 + 25 = 68% - 10-5 µm: 68 + 25 = 93% - 5-2 µm: 93 + 5 = 98% - 2-0 µm: 98 + 2 = 100% To recover 85%, the **cut size (d₅₀)** should be such that 85% (by weight) of the particles are larger than this size and are collected. From the cumulative: - At 68% (20 µm), next is 93% (5 µm). - 85% lies between 20-10 µm (68%) and 10-5 µm (93%). **Interpolate:** The fraction between 68% (20-10 µm) and 93% (10-5 µm) is: \[ \text{Fraction recovered from 10-20 μm} = 85\% - 68\% = 17\% \] The weight fraction in 10-5 μm is **25%**, so: \[ \text{Fraction of 10-5 μm needed} = \frac{17}{25} = 0.68 \] So, cut size is at: \[ d_{cut} = 10\, \mu m + (5\, \mu m - 10\, \mu m) \times 0.68 = 10\, \mu m - 3.4\, \mu m = 6.6\, \mu m \] But this is not a precise method; typically, the cut size is **the size at which 50% of the particles are collected** (d₅₀). Since we want 85% recovery, we need a cyclone with a cut size slightly less than 10 μm. For design, **use d₅₀ ≈ 10 μm** (since 68% is recovered above 10 μm, and with some collection below that, total reaches 85%). --- ### 4. **Properties of Air at 100°C (~373 K)** From your table (approximate): - **Density (ρg):** ~1.005 kg/m³ - **Viscosity (μ):** ~0.0314×10⁻³ Pa·s = 3.14×10⁻⁵ Pa·s --- ### 5. **Determine the Required Cut Size (d₅₀)** \[ d_{pc} \approx 10\, \mu m = 10 \times 10^{-6}\, m \] --- ### 6. **Cyclone Design Equation (Stairmand high-efficiency type)** The cut size for a cyclone is given approximately by: \[ d_{pc} = \sqrt{ \frac{9 \mu_g W_i}{2 \pi N_e V_g (\rho_p - \rho_g)} } \] Or, for a quick estimate (Perry’s Chemical Engineering Handbook): \[ d_{pc} = 0.0052 \left( \frac{\mu_g}{\rho_p - \rho_g} \right)^{0.5} \left( \frac{Q}{A} \right)^{0.5} \] Where: - \( d_{pc} \) = cut size (m) - \( Q \) = volumetric flow rate (m³/s) - \( \mu_g \) = gas viscosity (Pa·s) - \( \rho_p \) = particle density (kg/m³) - \( \rho_g \) = gas density (kg/m³) - \( A \) = cross-sectional area of cyclone (m²) But more commonly, the **optimum inlet velocity** for cyclones is about **15-20 m/s**. --- ### 7. **Determine Cyclone Diameter** The volumetric flow rate is: \[ Q = A \cdot V_{in} \] Assume \( V_{in} = 15 \) m/s. \[ A = \frac{Q}{V_{in}} = \frac{1.39}{15} = 0.0927\, m^2 \] For a rectangular slot inlet of width \( W \) and height \( h \), or for a circular cross section (if inlet is tangential): The **cyclone diameter** (D) is approximately: \[ A \approx D^2 \] \[ D = \sqrt{A} = \sqrt{0.0927} = 0.304\, m = 304\, mm \] So, a cyclone with **diameter of about 300 mm** is suitable for the flow. --- ### 8. **Estimate Original Solids Capacity (Loading Rate)** You ask for the "original capacity of opts (2g/(m²))" — I believe you want the **solid loading per area** or the **mass flux**. Assume a **solids loading** (c) in air (not provided, so let's assume a typical value or you can provide it). If **not given**, design is based on the flow and cyclone geometry. --- ### 9. **Summary Table** | Parameter | Value | |------------------------|--------------------| | Flow rate | 5000 m³/h (1.39 m³/s) | | Air density | 1.005 kg/m³ | | Air viscosity | 3.14×10⁻⁵ Pa·s | | Particle density | 2800 kg/m³ | | Required recovery | 85% | | Cut size (d₅₀) | ~10 μm | | Cyclone diameter | ~0.3 m (300 mm) | | Inlet velocity | 15 m/s | --- ### 10. **Conclusion & Recommendations** - **Use a cyclone separator of ~300 mm diameter** (or multiple cyclones in parallel for higher throughput). - **Inlet velocity**: 15 m/s - **Cut size**: ~10 μm, which will achieve ~85% solids recovery per your size distribution. - **If you need the exact mass loading per area (g/m²),** please provide the total solids feed rate (kg/h or g/s). --- #### **References** - Perry’s Chemical Engineering Handbook, 8th Edition, Section 20, Cyclone Design - Coulson & Richardson’s Chemical Engineering, Vol 2 --- If you have the **total dust loading in the air**, I can help you calculate the exact **g/(m²·s)** loading and suggest the number of cyclones needed for your plant! --- **If you clarify the meaning of "original capacity of opts (2g/(m²))", I can provide a more precise calculation.**