# Step-by-Step Solution: Filter Press Sizing for Downstream Filtration Let's break down the information and solve for the required number of filter presses. --- ## **Step 1: Calculate Fermentation Broth Volume** - **Tank diameter (D):** 1 m - **Tank height (H):** 2 m - **Loading factor:** 64% ### **Calculate the total tank volume:** \[ V_{\text{tank}} = \pi \left(\frac{D}{2}\right)^2 H = \pi (.5)^2 \times 2 = \pi \times .25 \times 2 \approx 1.57\, \text{m}^3 \] ### **Actual broth volume:** \[ V_{\text{broth}} = \text{Loading factor} \times V_{\text{tank}} = .64 \times 1.57 \approx 1.01\, \text{m}^3 = 101\, \text{L} \] --- ## **Step 2: Wet Filter Residue and Feed Volume** - **Wet filter residue to feed volume ratio = 30%** So, for every 1 L of feed, .3 L becomes wet residue. \[ V_{\text{residue}} = .3 \times V_{\text{broth}} = .3 \times 101 = 303\, \text{L} \] --- ## **Step 3: Filter Frame Filling Coefficient** - **Filling coefficient = .8** - **Total filter press frame volume (V):** 260 L (per press) Effective volume per cycle per press: \[ V_{\text{press, effective}} = .8 \times 260 = 208\, \text{L} \] --- ## **Step 4: Number of Filter Cycles Needed** Each press can handle 208 L of residue per cycle. \[ \text{Cycles required} = \frac{V_{\text{residue}}}{V_{\text{press, effective}}} = \frac{303}{208} \approx 1.46 \] But this is for one press and one batch's residue. --- ## **Step 5: Time Constraints** - **Total operation time allowed:** 20 hours - **Each filter press cycle time:** 2 hours Maximum number of cycles a single press can run in 20 hours: \[ \text{Cycles per press} = \frac{20}{2} = 10 \] --- ## **Step 6: Number of Presses Needed** Each press can process 208 L of residue per cycle, and can run 10 cycles in 20 hours: \[ \text{Total residue processed per press in 20 h} = 10 \times 208 = 208\, \text{L} \] But we only need to process 303 L of residue per batch (from the 101 L broth). \[ \text{Number of presses needed} = \frac{V_{\text{residue}}}{\text{Residue processed per press in 20 h}} = \frac{303}{208} \approx .146 \] Since you can't have a fraction of a filter press, **you only need 1 filter press** (since one press can handle much more than 303 L in 20 hours). --- ## **Final Answer** ### **Only 1 plate-and-frame filter press (BAS20/635-25) is required to filter one batch in 20 hours.** --- ### **Summary Table** | Parameter | Value | |---------------------------------|------------------| | Broth volume per batch | 101 L | | Wet filter residue per batch | 303 L | | Effective volume per press/cycle| 208 L | | Cycles per press in 20 h | 10 | | Residue handled per press/20 h | 208 L | | Number of presses needed | 1 | --- **If you have multiple fermenters or simultaneous batches, scale accordingly.**