1. **Problem Statement:** Calculate the theoretical volumetric flow rate and time to empty a water tank using given parameters in an Automated Water Management System (AWMS). 2. **Given Data and Assumptions:** - Tank volume $V_{\text{tank}} = 1000$ L $= 1$ m$^3$ - Pipe diameter $d = 0.0127$ m - Water head $h = 1$ m - Gravitational acceleration $g = 9.8$ m/s$^2$ - Discharge coefficient $C_d = 0.6014$ - Velocity coefficient $C_v = 0.97$ 3. **Step 1: Calculate Water Velocity at Tank Outlet** Using Torricelli's law with velocity coefficient: $$V_{wp} = C_v \sqrt{2gh}$$ Substitute values: $$V_{wp} = 0.97 \times \sqrt{2 \times 9.8 \times 1} = 0.97 \times \sqrt{19.6} \approx 0.97 \times 4.427 = 4.294 \text{ m/s}$$ 4. **Step 2: Calculate Cross-Sectional Area of Pipe** $$A = \pi \left( \frac{d}{2} \right)^2 = \pi \times (0.00635)^2 = \pi \times 4.032 \times 10^{-5} \approx 1.267 \times 10^{-4} \text{ m}^2$$ 5. **Step 3: Calculate Volumetric Flow Rate** Using discharge coefficient: $$Q = C_d \times A \times \sqrt{2gh}$$ Calculate $\sqrt{2gh}$: $$\sqrt{2 \times 9.8 \times 1} = 4.427$$ Then: $$Q = 0.6014 \times 1.267 \times 10^{-4} \times 4.427 \approx 3.374 \times 10^{-4} \text{ m}^3/\text{s}$$ Convert to liters per second: $$Q = 3.374 \times 10^{-4} \times 1000 = 0.3374 \text{ L/s}$$ Convert to cubic meters per hour: $$Q_w = Q \times 3600 = 3.374 \times 10^{-4} \times 3600 = 1.2146 \text{ m}^3/\text{h}$$ 6. **Step 4: Calculate Time to Empty the Tank** $$t = \frac{V_{\text{tank}}}{Q} = \frac{1}{3.374 \times 10^{-4}} \approx 2964 \text{ s}$$ Convert to minutes: $$t = \frac{2964}{60} \approx 49.4 \text{ minutes}$$ 7. **Step 5: General Flow Rate Expression** $$Q = C_d \left( \frac{\pi d^2}{4} \right) \sqrt{2gh}$$ This formula provides the baseline flow rate under normal conditions. 8. **Leakage Detection Criterion:** If the measured flow rate $Q_{\text{measured}}$ exceeds $110\%$ of the expected discharge $Q_{\text{threshold}}$ for a sustained time (e.g., 5-10 seconds), an alert for abnormal water flow is triggered. **Final answers:** - Volumetric flow rate $Q \approx 3.374 \times 10^{-4}$ m$^3$/s or 0.3374 L/s - Time to empty tank $t \approx 2964$ s or 49.4 minutes