Let's solve the problem step by step. **Given:** - Ore contains **5% copper sulphide** (C) and **95% gangue silicates** (G). - Rate constant for sulphide flotation, \( k_C = 0.2\, \text{min}^{-1} \) - Rate constant for gangue flotation, \( k_G = 0.01\, \text{min}^{-1} \) - Total ore treated per hour = **100 t/h** - Number of cells = **10** - Total residence time in bank = **20 min** Let's answer each part: --- ## (i) **What grade of concentrate would be achieved in a bank of 10 cells with 20 min residence time?** **Step 1: Fraction floated (Recovery) for each component** For **first-order kinetics** in a plug flow (series of tanks), the fraction recovered is: \[ \text{Fraction floated (recovered)} = 1 - e^{-kt} \] where \( k \) is the flotation rate constant and \( t \) is residence time. So, - For **copper sulphide** (\( k_C = 0.2 \)): \[ R_C = 1 - e^{-0.2 \times 20} = 1 - e^{-4} \] \[ e^{-4} \approx 0.0183 \] \[ R_C = 1 - 0.0183 = 0.9817 \text{ or } 98.17\% \] - For **gangue** (\( k_G = 0.01 \)): \[ R_G = 1 - e^{-0.01 \times 20} = 1 - e^{-0.2} \] \[ e^{-0.2} \approx 0.8187 \] \[ R_G = 1 - 0.8187 = 0.1813 \text{ or } 18.13\% \] --- **Step 2: Calculate amount of each in feed** Suppose **100 t/h** feed: - Copper sulphide: \( 0.05 \times 100 = 5\, \text{t/h} \) - Gangue: \( 0.95 \times 100 = 95\, \text{t/h} \) --- **Step 3: Calculate amounts reporting to concentrate** - **Copper sulphide in concentrate:** \[ = \text{Feed} \times \text{Fraction recovered} = 5 \times 0.9817 = 4.9085\, \text{t/h} \] - **Gangue in concentrate:** \[ = 95 \times 0.1813 = 17.2235\, \text{t/h} \] --- **Step 4: Total concentrate produced** \[ \text{Total concentrate} = 4.9085 + 17.2235 = 22.132\, \text{t/h} \] --- **Step 5: Grade of concentrate** \[ \text{Grade} = \frac{\text{Copper sulphide in concentrate}}{\text{Total concentrate}} \times 100\% \] \[ \text{Grade} = \frac{4.9085}{22.132} \times 100\% = 22.18\% \] --- ## (ii) **What would be the grade after 20 min?** This is the same as above, since the residence time is 20 min for the bank. The grade after 20 min is the grade of the concentrate calculated above. --- ## **Final Answers** ### (i) The grade of concentrate achieved in a bank of 10 cells (20 min residence time) is: \[ \boxed{22.2\%} \] ### (ii) The grade after 20 min (i.e., at the end of the bank) is: \[ \boxed{22.2\%} \] --- **Step-by-step summary:** 1. Calculated recoveries using first-order kinetics. 2. Determined mass of each component in concentrate. 3. Found total concentrate mass. 4. Calculated grade as (mass of copper sulphide in concentrate) / (total concentrate) × 100%. --- **If you need the calculation shown with more decimal places or a different format, let me know!**