1. **Problem Statement:** Calculate the water flow rate $Q_w$ through a pipe leakage of diameter 0.5 inch (12.7 mm) from a 1000-liter water tank and determine the time to empty the tank. 2. **Given Data:** - Water tank volume = 1000 liters = 1 m³ - Pipe diameter $d = 0.5$ inch = 0.0127 m - Water velocity coefficient $C_{wv} = 0.97$ - Gravity acceleration $g = 9.8$ m/s² - Height of water tank $h = 1$ m - Pipe discharge coefficient $C_{df} = 0.6014$ 3. **Step 1: Calculate water velocity $V_{wp}$ in the pipe** Using the formula: $$V_{wp} = C_{wv} \sqrt{2gh}$$ Substitute values: $$V_{wp} = 0.97 \times \sqrt{2 \times 9.8 \times 1} = 0.97 \times \sqrt{19.6} = 0.97 \times 4.427 = 4.2843\, m/s$$ 4. **Step 2: Calculate flow area $F$ of the pipe** $$F = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = 3.142 \times (0.0127/2)^2 = 3.142 \times 0.00635^2 = 3.142 \times 0.00004032 = 0.0001267\, m^2$$ 5. **Step 3: Calculate water flow volume $V_{wf}$ from the pipe** Using: $$V_{wf} = C_{df} \times F \times V_{wp}$$ Substitute values: $$V_{wf} = 0.6014 \times 0.0001267 \times 4.2843 = 0.0003275\, m^3/s$$ 6. **Step 4: Convert flow rate to $m^3/h$** $$Q_w = V_{wf} \times 3600 = 0.0003275 \times 3600 = 1.179\, m^3/h$$ 7. **Step 5: Calculate time to empty the tank** Tank volume = 1 $m^3$ Time $t$ in hours: $$t = \frac{\text{Tank volume}}{Q_w} = \frac{1}{1.179} = 0.848\, h$$ Convert to seconds: $$t = 0.848 \times 3600 = 3053\, seconds$$ Convert to minutes: $$t = \frac{3053}{60} = 50.88\, minutes$$ **Note:** The original problem's flow area calculation used $F=3.142 \times 0.0127 \times 0.0127 = 0.00056064$ which is incorrect for area (should be radius squared). Using radius squared gives the correct flow area. **Final answer:** The water flow rate through the pipe leakage is approximately $1.179\, m^3/h$, and it will take about 50.88 minutes to empty the 1000-liter tank through the 0.5 inch pipe leakage. This model helps detect leakage by monitoring if the flow rate exceeds a threshold, triggering a sensor alert.