Let's break down and solve the flotation problem step by step: ### **Given Data** - **Ore composition:** 5% copper sulphide, 95% gangue silicates - **Feed rate (F):** 100 tons/hour - **Rate constant for sulphide (k₁):** 0.2 min⁻¹ - **Rate constant for gangue (k₂):** 0.01 min⁻¹ - **Number of cells:** 10 - **Total residence time:** 20 min We assume first-order kinetics for flotation. --- ## **Step 1: Calculate Fraction Recovered for Each Component** For a bank of cells operating as **perfect mixers in series** (i.e., plug flow), the recovery \( R \) for a first-order process with rate constant \( k \) and total residence time \( t \) is: \[ R = 1 - e^{-kt} \] ### **A. Copper Sulphide (k₁ = 0.2 min⁻¹):** \[ R_{\text{CuS}} = 1 - e^{-0.2 \times 20} = 1 - e^{-4} \] \[ e^{-4} \approx 0.0183 \] \[ R_{\text{CuS}} = 1 - 0.0183 = 0.9817 \quad \text{or} \quad 98.17\% \] ### **B. Gangue Silicates (k₂ = 0.01 min⁻¹):** \[ R_{\text{Gangue}} = 1 - e^{-0.01 \times 20} = 1 - e^{-0.2} \] \[ e^{-0.2} \approx 0.8187 \] \[ R_{\text{Gangue}} = 1 - 0.8187 = 0.1813 \quad \text{or} \quad 18.13\% \] --- ## **Step 2: Calculate Amounts Floated (in Concentrate)** ### **Feed composition in 100 tons:** - Copper sulphide: \( 0.05 \times 100 = 5 \) tons - Gangue silicates: \( 0.95 \times 100 = 95 \) tons ### **A. Copper sulphide in concentrate:** \[ \text{CuS in concentrate} = 5 \times 0.9817 = 4.9085 \text{ tons} \] ### **B. Gangue in concentrate:** \[ \text{Gangue in concentrate} = 95 \times 0.1813 = 17.2235 \text{ tons} \] ### **Total concentrate produced:** \[ \text{Total concentrate} = 4.9085 + 17.2235 = 22.132 \text{ tons} \] --- ## **Step 3: Calculate Grade of Concentrate** Grade (\%) of copper sulphide in the concentrate: \[ \text{Grade} = \frac{\text{CuS in concentrate}}{\text{Total concentrate}} \times 100 \] \[ \text{Grade} = \frac{4.9085}{22.132} \times 100 = 22.18\% \] --- ## **Step 4: Final Answers** ### **(i) What grade of concentrate would be achieved in a bank of 10 cells treating 100 t/h if the total residence time is 20 min?** \[ \boxed{22.2\% \text{ copper sulphide (rounded to 1 decimal place)}} \] --- ### **(ii) What would be the grade after 20 min?** Since the total residence time is 20 min, the answer is the same as above: \[ \boxed{22.2\% \text{ copper sulphide}} \] --- ### **Summary Table** | | Copper sulphide | Gangue silicates | Total | |---|-----------------|------------------|-------| | Feed (t) | 5 | 95 | 100 | | Floated (t) | 4.91 | 17.22 | 22.13 | | Grade (%) | **22.2** | 77.8 | 100 | --- ## **Conclusion** - **(i) The grade of the concentrate after 20 minutes in 10 cells is \( \boxed{22.2\%} \) copper sulphide.** - **(ii) After 20 minutes, the grade remains \( \boxed{22.2\%} \) copper sulphide.** **Let me know if you want recovery calculations or further breakdown!**

Let's carefully analyze and solve the flotation problem step-by-step. --- ### **Given Data:** - **Ore composition:** - Copper sulphide (CuS): 5% - Gangue silicates: 95% - **Feed rate (F):** 100 tons/hour - **Rate constants:** - For copper sulphide, \(k_1 = 0.2 \text{ min}^{-1}\) - For gangue silicates, \(k_2 = 0.01 \text{ min}^{-1}\) - **Number of cells in the bank:** 10 - **Total residence time in the bank:** 20 minutes --- ### **Step 1: Understand the flotation kinetics** Assuming first-order kinetics for flotation: \[ R = 1 - e^{-k t} \] where: - \( R \) = recovery fraction - \( k \) = rate constant - \( t \) = residence time --- ### **Step 2: Calculate the recovery for each component** **A. Copper sulphide:** \[ R_1 = 1 - e^{-0.2 \times 20} = 1 - e^{-4} \] Calculate \( e^{-4} \): \[ e^{-4} \approx 0.0183 \] So, \[ R_1 = 1 - 0.0183 = 0.9817 \quad \text{or} \quad 98.17\% \] **B. Gangue silicates:** \[ R_2 = 1 - e^{-0.01 \times 20} = 1 - e^{-0.2} \] Calculate \( e^{-0.2} \): \[ e^{-0.2} \approx 0.8187 \] So, \[ R_2 = 1 - 0.8187 = 0.1813 \quad \text{or} \quad 18.13\% \] --- ### **Step 3: Calculate the amount of each component in the concentrate** **Initial feed:** - Copper sulphide: \(5\, \text{tons}\) (since 5% of 100 tons) - Gangue: \(95\, \text{tons}\) **In the concentrate:** - Copper sulphide: \[ \text{CuS in concentrate} = 5 \times R_1 = 5 \times 0.9817 = 4.9085\, \text{tons} \] - Gangue: \[ \text{Gangue in concentrate} = 95 \times R_2 = 95 \times 0.1813 = 17.2235\, \text{tons} \] **Total concentrate:** \[ 4.9085 + 17.2235 = 22.132\, \text{tons} \] --- ### **Step 4: Calculate the grade of copper sulphide in the concentrate** \[ \text{Grade (CuS)} = \frac{4.9085}{22.132} \times 100 \approx 22.2\% \] --- ### **Final answers:** #### **(i) Grade of concentrate after 20 minutes:** \[ \boxed{\approx 22.2\% \text{ CuS}} \] --- #### **(ii) Grade after 20 minutes:** Since the total residence time is 20 minutes, the grade remains approximately **22.2%**. --- ### **Summary:** | Parameter | Value | |--------------|---------| | Recovery of CuS | 98.2% | | Recovery of Gangue | 18.1% | | Concentrate grade | **22.2% CuS** | --- **Note:** This calculation assumes ideal first-order kinetics and perfect mixing in each cell. Let me know if you'd like to explore recovery percentages or other parameters!