Let's solve the problem step by step. ## **Problem Restatement** - Feed with particles of three sizes: - 25% of **50 μm** radius, **conversion time = 5 min** - 20% of **100 μm** radius, **conversion time = 10 min** - 55% of **200 μm** radius, **conversion time = 20 min** - Residence time in reactor = **8 min** - Find **overall conversion** of solids. --- ## **Step 1: Conversion for Each Particle Size** For **first-order kinetics** (or constant-rate process), **fractional conversion** (\( X \)) for a given residence time (\( t \)) and complete conversion time (\( t_{c} \)) is: \[ X = \frac{t}{t_c} \quad \text{if } t < t_c \] \[ X = 1 \quad \text{if } t \geq t_c \] But, generally, for solid conversion in a reactor, if a particle needs time \( t_c \) for complete conversion, then after time \( t \): \[ \text{Fraction unconverted} = 1 - \frac{t}{t_c} \quad (t < t_c) \] \[ \text{Fraction converted} = \frac{t}{t_c} \quad (t < t_c) \] \[ \text{Fraction converted} = 1 \quad (t \geq t_c) \] --- ### **50 μm Particles** - Fraction of feed: \( w_1 = .25 \) - Complete conversion time: \( t_{c1} = 5 \) min - Residence time: \( t = 8 \) min \[ X_1 = \min\left(1, \frac{8}{5}\right) = 1 \] --- ### **100 μm Particles** - Fraction of feed: \( w_2 = .20 \) - Complete conversion time: \( t_{c2} = 10 \) min \[ X_2 = \frac{8}{10} = .8 \] --- ### **200 μm Particles** - Fraction of feed: \( w_3 = .55 \) - Complete conversion time: \( t_{c3} = 20 \) min \[ X_3 = \frac{8}{20} = .4 \] --- ## **Step 2: Weighted Average Overall Conversion** \[ X_{\text{overall}} = w_1 X_1 + w_2 X_2 + w_3 X_3 \] Plug values: \[ X_{\text{overall}} = (.25 \times 1) + (.20 \times .8) + (.55 \times .4) \] \[ = .25 + .16 + .22 = .63 \] --- ## **Final Answer** \[ \boxed{.63} \] **The conversion of solids on the grate for a residence time of 8 min in the reactor is .63, or 63%.**