Let's break down the problem and solve each part step-by-step. **Given:** - **Gas flow rate:** \(3000 \, \text{m}^3/\text{min}\) - **Dust concentration:** \(2.00 \, \text{g}/\text{m}^3\) - **Filter bag size:** Length \(= 14 \, \text{ft}\), Diameter \(= 10 \, \text{in}\) - **Work in English units.** We'll need to make some standard assumptions for baghouse design (e.g., typical filter velocity, number of compartments, etc.), unless otherwise stated. --- ### Step 1: Convert Flow Rate to ft³/min \[ 1 \, \text{m}^3 = 35.315 \, \text{ft}^3 \] \[ Q = 3000 \, \text{m}^3/\text{min} \times 35.315 \, \text{ft}^3/\text{m}^3 = 105,945 \, \text{ft}^3/\text{min} \] --- ### Step 2: Bag Area Calculation #### **Bag Dimensions in Feet:** - **Diameter:** \(10 \, \text{in} = \frac{10}{12} \, \text{ft} = 0.833 \, \text{ft}\) - **Length:** \(14 \, \text{ft}\) #### **Surface Area of One Bag (cylinder, neglecting ends):** \[ A_{\text{bag}} = \pi \times \text{diameter} \times \text{length} \] \[ A_{\text{bag}} = \pi \times 0.833 \, \text{ft} \times 14 \, \text{ft} \approx 36.6 \, \text{ft}^2 \] --- ### Step 3: Cake Filtering Velocity, \(v\) (ft/min) Typical design velocities (air-to-cloth ratio) for reverse-air baghouses are **2–3 ft/min**. Let's use \(v = 2.5 \, \text{ft/min}\) (typical value). --- ### Step 4: Total Cloth Area Needed \[ \text{Cloth Area} = \frac{\text{Flow Rate}}{\text{Filtering Velocity}} \] \[ \text{Cloth Area} = \frac{105,945 \, \text{ft}^3/\text{min}}{2.5 \, \text{ft}/\text{min}} = 42,378 \, \text{ft}^2 \] --- ### Step 5: Number of Bags \[ \text{Number of Bags} = \frac{\text{Total Cloth Area}}{\text{Area per Bag}} \] \[ \text{Number of Bags} = \frac{42,378 \, \text{ft}^2}{36.6 \, \text{ft}^2} \approx 1,158 \text{ bags} \] --- ### Step 6: Number of Compartments Typical baghouses have 6–10 compartments. Let's use **8 compartments** (common for cleaning cycles). --- ### Step 7: Cloth Area per Compartment \[ \text{Cloth Area per Compartment} = \frac{42,378 \, \text{ft}^2}{8} = 5,297 \, \text{ft}^2 \] --- ### Step 8: Bags per Compartment \[ \text{Bags per Compartment} = \frac{5,297 \, \text{ft}^2}{36.6 \, \text{ft}^2} \approx 145 \text{ bags} \] --- ## **Final Answers:** (a) **Cake filtering velocity:**   \(\boxed{2.5 \, \text{ft/min}}\) *(typical design value)* (b) **Number of compartments:**   \(\boxed{8}\) (c) **Cloth area per compartment:**   \(\boxed{5,297 \, \text{ft}^2}\) (d) **Filtering bags per compartment:**   \(\boxed{145}\) --- **If you need more or less conservative design numbers, adjust the filtering velocity and recalculate accordingly.**