# Water Treatment Plant Calculation Given: - **Plant flow:** \( Q_{\text{total}} = 50{,}000 \) m³/day - **Side stream flow:** \( Q_{\text{side}} = 2{,}000 \) m³/day - **Alum dose:** \( C_{\text{alum}} = 1.5 \) g/L - **Rapid mix tank detention time:** \( t = 45 \) s - **Velocity gradient:** \( G = 450 \) s\(^{-1}\) - **Temperature:** \( 15^\circ \)C - **Impeller:** Flat-blade - **Alum:** \( \text{Al}_2(\text{SO}_4)_3 \) - **Water density:** \( \rho \approx 100 \) kg/m³ - **Viscosity at 15°C:** \( \mu \approx 1.14 \times 10^{-3} \) Pa·s --- ## (a) Alum Concentration in Tank (mg/L) **Alum added to side stream:** - \( 1.5 \) g/L in \( 2,000 \) m³/day **Convert m³ to L:** \[ 2,000~\text{m}^3 = 2,000,000~\text{L} \] **Total alum mass per day:** \[ \text{Total~alum~added} = 1.5~\text{g/L} \times 2,000,000~\text{L} = 3,000,000~\text{g} = 3,000~\text{kg} \] **After mixing with total flow (assuming complete mixing):** \[ \text{Final concentration in main flow} = \frac{3,000~\text{kg}}{50,000~\text{m}^3} \] \[ = \frac{3,000,000~\text{g}}{50,000,000~\text{L}} = .06~\text{g/L} = 60~\text{mg/L} \] **Final answer:** **Alum concentration in the main flow after mixing: \(\boxed{60~\text{mg/L}}\)** --- ## (b) Daily Alum Mass (kg/day) Already calculated above: \[ \boxed{3,000~\text{kg/day}} \] --- ## (c) Tank Dimensions and Impeller Diameter ### **Step 1: Tank Volume** The rapid mix tank receives the **total plant flow**. \[ Q = 50,000~\text{m}^3/\text{day} = \frac{50,000}{24 \times 360}~\text{m}^3/\text{s} \] \[ = \frac{50,000}{86,400}~\text{m}^3/\text{s} \approx .5787~\text{m}^3/\text{s} \] **Detention time:** \[ t = 45~\text{s} \] **Tank volume:** \[ V = Q \times t = .5787~\text{m}^3/\text{s} \times 45~\text{s} \approx 26.~\text{m}^3 \] ### **Step 2: Tank Dimensions** Assume **cubic tank:** Let side = \( a \), so \( V = a^3 \). \[ a = V^{1/3} = (26.)^{1/3} \approx 2.96~\text{m} \] **Tank dimensions:** \[ \boxed{2.96~\text{m} \times 2.96~\text{m} \times 2.96~\text{m}} \] ### **Step 3: Impeller Diameter** For rapid mix, a **typical ratio** is: \[ \frac{D}{T} = .3 \text{ to } .5 \] where \( D \) = impeller diameter, \( T \) = tank width. Let’s use \( D/T = .33 \): \[ D = .33 \times 2.96~\text{m} \approx .98~\text{m} \] \[ \boxed{.98~\text{m}~(\text{impeller diameter})} \] --- ## (d) Power Input (kW) **Given:** \[ G = 450~\text{s}^{-1},\ V = 26.~\text{m}^3,\ \mu = 1.14 \times 10^{-3}~\text{Pa}\cdot \text{s} \] \[ G = \sqrt{\frac{P}{\mu V}} \] \[ P = G^2 \times \mu \times V \] \[ = (450)^2 \times 1.14 \times 10^{-3} \times 26. \] \[ = 202,500 \times .00114 \times 26. \] \[ = 230.85 \times 26. \] \[ = 6,002~\text{W} = 6.~\text{kW} \] --- ## **Summary Table** | Parameter | Value | |------------------------------- |--------------------------------| | (a) Alum concentration | **60 mg/L** | | (b) Daily alum mass | **3,000 kg/day** | | (c) Tank dimensions (cube) | **2.96 m × 2.96 m × 2.96 m** | | (c) Impeller diameter | **.98 m** | | (d) Power input | **6. kW** |