Let's solve each part step by step: --- ## **Given Data** - **Filter area:** \( A = 1.8\, \text{cm}^2 = 1.8 \times 10^{-4}\, \text{m}^2 \) - **Solids per ml of filtrate:** \( C = .25\, \text{g/ml} = 250\, \text{kg/m}^3 \) - **Filtrate viscosity:** \( \mu = 1.38\, \text{cP} = 1.38 \times 10^{-3}\, \text{Pa} \cdot \text{s} \) - **Filtrate volumes and times:** From the table - **Pressure drops:** From the table --- ## **a) Evaluate the specific cake resistance (\(\alpha\)) as a function of pressure** ### **Filtration Equation** The constant pressure filtration equation (for incompressible cake): \[ \frac{t}{V} = \frac{\mu \alpha C}{2 A^2 \Delta P} V + \frac{\mu R_m}{A \Delta P} \] Where: - \( t \) = time (s) - \( V \) = filtrate volume (m³) - \( \mu \) = viscosity (Pa·s) - \( \alpha \) = specific cake resistance (m/kg) - \( C \) = mass of solids per volume of filtrate (kg/m³) - \( A \) = filter area (m²) - \( \Delta P \) = pressure drop (Pa) - \( R_m \) = medium resistance ### **Step 1: Convert Units** - \( A = 1.8\, \text{cm}^2 = 1.8 \times 10^{-4}\, \text{m}^2 \) - \( C = .25\, \text{g/ml} = .25 \times 100 = 250\, \text{kg/m}^3 \) - \( \mu = 1.38\, \text{cP} = 1.38 \times 10^{-3}\, \text{Pa} \cdot \text{s} \) - Pressure (kPa to Pa): \( 1\, \text{kPa} = 100\, \text{Pa} \) - Volume (ml to m³): \( 1\, \text{ml} = 1 \times 10^{-6}\, \text{m}^3 \) ### **Step 2: Plot \( t \) vs \( V \) and Find Slope** For each pressure, plot \( t \) vs \( V \) (convert \( V \) to m³), fit a straight line. The slope \( S \) is: \[ S = \frac{dt}{dV} \] ### **Let’s calculate for one pressure (e.g., 33.3 kPa):** - \( V \) (ml): 10, 15, 20, 25, 30, 35, 40, 45, 50 - \( t \) (s): 12, 26, 49, 75, 110, 149, 193, 240, 301 Convert \( V \) to m³: - 10 ml = \( 1. \times 10^{-5} \) m³ - 50 ml = \( 5. \times 10^{-5} \) m³ Estimate slope (\( S \)) between first and last points: \[ S = \frac{301 - 12}{5. \times 10^{-5} - 1. \times 10^{-5}} = \frac{289}{4. \times 10^{-5}} = 7.225 \times 10^6 \, \text{s/m}^3 \] ### **Step 3: Calculate \(\alpha\)** From the filtration equation: \[ S = \frac{\mu \alpha C}{A^2 \Delta P} V + \frac{\mu R_m}{A \Delta P} \] At large \( V \), the intercept is small compared to the slope term, so: \[ \frac{dt}{dV} \approx \frac{\mu \alpha C}{A^2 \Delta P} V \] But for the initial slope, rearrange for \(\alpha\): \[ \alpha = \frac{S A^2 \Delta P}{\mu C V} \] - \( S = 7.225 \times 10^6 \) s/m³ - \( A = 1.8 \times 10^{-4} \) m² - \( \Delta P = 33.3 \times 10^3 \) Pa - \( \mu = 1.38 \times 10^{-3} \) Pa·s - \( C = 250 \) kg/m³ - \( V = 5. \times 10^{-5} \) m³ (use largest volume) Plug in values: \[ \alpha = \frac{7.225 \times 10^6 \cdot (1.8 \times 10^{-4})^2 \cdot 33.3 \times 10^3}{1.38 \times 10^{-3} \cdot 250 \cdot 5. \times 10^{-5}} \] Calculate step by step: - \( (1.8 \times 10^{-4})^2 = 3.24 \times 10^{-8} \) - Numerator: \( 7.225 \times 10^6 \cdot 3.24 \times 10^{-8} \cdot 33.3 \times 10^3 \) - \( 7.225 \times 10^6 \cdot 3.24 \times 10^{-8} = .234 \) - \( .234 \cdot 33.3 \times 10^3 = .234 \cdot 33300 = 7792.2 \) - Denominator: \( 1.38 \times 10^{-3} \cdot 250 \cdot 5. \times 10^{-5} \) - \( 1.38 \times 10^{-3} \cdot 250 = .345 \) - \( .345 \cdot 5. \times 10^{-5} = 1.725 \times 10^{-5} \) - \( \alpha = \frac{7792.2}{1.725 \times 10^{-5}} = 4.52 \times 10^8 \) m/kg **Repeat for other pressures.** This gives \(\alpha\) as a function of pressure. --- ## **b) Determine the compressibility** The relationship is: \[ \alpha = \alpha_ \left( \frac{\Delta P}{\Delta P_} \right)^s \] Take log on both sides: \[ \log \alpha = \log \alpha_ + s \log \Delta P \] Plot \(\log \alpha\) vs \(\log \Delta P\). The slope gives compressibility \(s\). - If \(s \approx \): incompressible cake. - If \(s \approx 1\): highly compressible. **Use calculated \(\alpha\) at each \(\Delta P\) to find \(s\).** --- ## **c) Calculate required pressure drop for scale-up** Given: - Filter area (\(A\)) = 15 m² - Culture to process (\(V\)) = 20 m³ in 1 hr (\(= 360\) s) - \(C = 250\, \text{kg/m}^3\) - \(\mu = 1.38 \times 10^{-3}\) Pa·s Use the filtration equation: \[ t = \frac{\mu \alpha C}{2 A^2 \Delta P} V^2 \] Rearrange for \(\Delta P\): \[ \Delta P = \frac{\mu \alpha C V^2}{2 A^2 t} \] Use average \(\alpha\) (from above, say \(4.52 \times 10^8\) m/kg). Plug values: - \(V = 20\) m³ - \(A = 15\) m² - \(t = 360\) s \[ \Delta P = \frac{1.38 \times 10^{-3} \times 4.52 \times 10^8 \times 250 \times (20)^2}{2 \times (15)^2 \times 360} \] Calculate step by step: - \(1.38 \times 10^{-3} \times 4.52 \times 10^8 = 6.24 \times 10^5\) - \(6.24 \times 10^5 \times 250 = 1.56 \times 10^8\) - \((20)^2 = 400\) - Numerator: \(1.56 \times 10^8 \times 400 = 6.24 \times 10^{10}\) - Denominator: \(2 \times 225 \times 360 = 450 \times 360 = 1,620,000\) \[ \Delta P = \frac{6.24 \times 10^{10}}{1.62 \times 10^6} = 38,518\, \text{Pa} = 38.5\, \text{kPa} \] --- ## **Summary of Answers** ### **a) Specific cake resistance:** - Calculated values for each pressure (see above; e.g., \(4.52 \times 10^8\) m/kg at 33.3 kPa). ### **b) Compressibility:** - Calculate \(s\) from the log-log plot of \(\alpha\) vs \(\Delta P\). ### **c) Required pressure drop for scale-up:** - **Required \(\Delta P \approx 38.5\, \text{kPa}\)** --- ### **If you need the full calculation table for each pressure, or plotting log-log for compressibility, let me know!**