Let's break the problem into its parts and solve step by step. --- ## **Given Data** - **Slurry concentration**: 18 g/L CaCO₃ in water - **μ (water viscosity)**: 1 cp = \( 1 \times 10^{-3} \) Pa·s - **α₀ (initial specific cake resistance)**: \( 1.2 \times 10^{12} \) m/kg - **Frame size**: 1 square frame, 30 cm side = .3 m - **ΔP, slope, intercept:** | ΔP (N/cm²) | slope | intercept | |------------|-------|-----------| | 6 | 25 | 26.5 | | 36 | 7 | 6.9 | - \( V \) is in **liters** --- ## **Background** The general filtration equation using a plate and frame filter press is: \[ \frac{dt}{dV} = \frac{\mu}{A^2 \Delta P} \left[ \alpha_ C_s V + R_m \right] \] - \( \frac{dt}{dV} \): Slope from the filtration plot - \( \mu \): Viscosity (kg/m·s) - \( A \): Total filter area (m²) - \( \Delta P \): Pressure drop (N/m²) - \( \alpha_ \): Specific cake resistance (m/kg) - \( C_s \): Mass of solids per volume of filtrate (kg/m³) - \( R_m \): Resistance of filter medium The plot of \( \frac{dt}{dV} \) vs \( V \) gives: - Slope = \( \frac{\mu \alpha_ C_s}{A^2 \Delta P} \) - Intercept = \( \frac{\mu R_m}{A^2 \Delta P} \) --- ## **Step 1: Convert Units and Data** ### **Pressure** - 1 N/cm² = \( 10^4 \) N/m² (Pa) - ΔP₁ = 6 N/cm² = \( 6 \times 10^4 \) Pa - ΔP₂ = 36 N/cm² = \( 36 \times 10^4 \) Pa ### **Concentration** \( C_s = 18 \) g/L = \( .018 \) kg/L = \( 18 \) kg/m³ ### **Slope Units** - Slope is in min/L (from \( dt/dV \) vs \( V \), with \( V \) in L and \( t \) in min). - Convert to seconds and cubic meters as needed for SI consistency. --- ## **Step 2: Write Slope Equations** \[ \text{Slope} = \frac{\mu \alpha_ C_s}{A^2 \Delta P} \] For both data points: ### **At ΔP = 6 N/cm²:** \[ 25 = \frac{(1 \times 10^{-3}) \times (1.2 \times 10^{12}) \times 18}{A^2 \times 6 \times 10^4} \] ### **At ΔP = 36 N/cm²:** \[ 7 = \frac{(1 \times 10^{-3}) \times (1.2 \times 10^{12}) \times 18}{A^2 \times 36 \times 10^4} \] --- ## **Step 3: Solve for A (Total Area)** Let’s use the equation for ΔP = 6 N/cm². ### **First, resolve units:** - Slope needs to be in s/m³ (since SI units): \[ 25~(\text{min}/\text{L}) = 25 \times 60~\text{s}/100~\text{m}^3 = 1.5~\text{s}/\text{m}^3 \] ### **Plug values:** \[ 1.5 = \frac{(1 \times 10^{-3}) \times (1.2 \times 10^{12}) \times 18}{A^2 \times 6 \times 10^4} \] Calculate numerator: \[ (1 \times 10^{-3}) \times (1.2 \times 10^{12}) \times 18 = 1.2 \times 18 \times 10^9 = 21.6 \times 10^9 = 2.16 \times 10^{10} \] So, \[ 1.5 = \frac{2.16 \times 10^{10}}{A^2 \times 6 \times 10^4} \] \[ A^2 = \frac{2.16 \times 10^{10}}{1.5 \times 6 \times 10^4} = \frac{2.16 \times 10^{10}}{9 \times 10^4} = \frac{2.16}{9} \times 10^{10-4} = .24 \times 10^6 = 2.4 \times 10^5 \] \[ A = \sqrt{2.4 \times 10^5} = \sqrt{2.4} \times 10^{2.5} \] \[ \sqrt{2.4} \approx 1.549 \] \[ 10^{2.5} = 10^2 \times 10^{.5} = 100 \times 3.16 = 316 \] \[ A = 1.549 \times 316 \approx 489.5~\text{m}^2 \] --- ## **Step 4: Number of Plates** - Each plate is a square of 30 cm = .3 m per side. - Area per plate = \( .3 \times .3 = .09~\text{m}^2 \) \[ \text{Number of plates} = \frac{A}{A_{\text{plate}}} = \frac{489.5}{.09} \approx 5449 \] --- ## **Final Answers** ### **(a) Total Area of Filtration** \[ \boxed{489.5~\text{m}^2} \] ### **(b) Total Number of Plates** \[ \boxed{5449} \] --- ### **Summary Table** | Part | Quantity | Value | Units | |------|----------|--------------|--------------| | (a) | Area | 489.5 | m² | | (b) | Plates | 5449 | plates | --- If you have any follow-up or want the calculation using the other data point or for intercept, let me know!