# Problem 1: Fire Flow and Duration for Two Adj Buildings ## Given Data: ### Building 1 (Wooden Frame) - Number of floors: 4 - Area per floor: 509 m² Building 2 (Ordinary Construction) - Number of floors: 5 - Area per floor: 900 m² --- ## Step 1: Calculate the fire flow for each building ### Fire flow formula (assuming the **National Building Fire Use (NBFU)** method): \[ Q = \text{Area} \times \text{Fire flow factor} \] *Note: For typical calculations, the fire flow factor varies based on construction type. For simplicity, assume standard factors:* - Wooden buildings: **.2 L/s per m²** - Ordinary construction: **.4 L/s per m²** --- ### Fire flow for Building 1 (Wooden Frame): \[ Q_1 = 509 \text{ m}^2 \times 4 \text{ floors} \times .2 \text{ L/s per m}^2 \] \[ Q_1 = 509 \times 4 \times .2 = 407.2 \text{ L/s} \] --- ### Fire flow for Building 2 (Ordinary Construction): \[ Q_2 = 900 \text{ m}^2 \times 5 \text{ floors} \times .4 \text{ L/s per m}^2 \] \[ Q_2 = 900 \times 5 \times .4 = 180 \text{ L/s} \] --- ## Step 2: Determine fire flow for both buildings connected - When connected, the **total fire flow** is the **sum** of individual flows: \[ Q_{total} = Q_1 + Q_2 = 407.2 + 180 = \boxed{2207.2 \text{ L/s}} \] --- ## Step 3: Estimate fire duration - Assume a typical fire duration of **2 hours** (or 720 seconds). - The **total water requirement**: \[ W = Q_{total} \times \text{duration} \] \[ W = 2207.2 \text{ L/s} \times 720 \text{ s} = 15,887,040 \text{ L} \] --- # **Final Summary for Problem 1:** | Quantity | Building 1 | Building 2 | Both Connected | |---|---|---|---| | Fire flow (L/s) | 407.2 | 180 | 2207.2 | | Fire duration | 2 hours | 2 hours | 2 hours | | Total water (L) | 2,927,424 | 12,959,616 | 15,887,040 | --- # Problem 2: Water Requirements for a Community ## Given Data: - Population at design year: **1,200,000** - Municipal water demand: **610 Lpd (Liters per capita per day)** - Community population: **1,200,000** --- ## Step 1: Calculate total water demand \[ Q_{demand} = \text{Population} \times \text{Water demand per capita} \] \[ Q_{demand} = 1,200,000 \times 610 \text{ Lpd} \] \[ Q_{demand} = 732,000,000 \text{ L} \] --- ## Step 2: Convert to **L/sec** for design capacity \[ Q_{sec} = \frac{732,000,000}{24 \times 360} = \frac{732,000,000}{86,400} \approx 8,472 \text{ L/sec} \] --- ## Step 3: Fire flow estimation using **NBFU formula** \[ Q_f = .2 \times \sqrt{A} \] Where \(A\) is the **area** of the community or a representative zone. Since no specific area is given, assume the **fire flow** is proportional to the total water demand. **Alternatively**, based on standard practice, a **fire flow** of at least **3,600 L/min (60 L/sec)** is often used for large communities. --- ## Step 4: Design capacity of water treatment plant - To meet **daily demand**: \[ Q_{plant} = 8,472 \text{ L/sec} \] - **Design capacity** should include a **peak factor**, typically 2: \[ Q_{design} = 2 \times 8,472 = 16,944 \text{ L/sec} \] --- ## Step 5: Design capacity of water distribution system - The **distribution system capacity** should accommodate both **average demand** and **fire flows**. - Given the **fire flow** estimated at **3,600 L/min (60 L/sec)**, the **system capacity**: \[ Q_{system} = \max(Q_{demand}, Q_{fire}) \] \[ Q_{system} \approx \max(8,472 \text{ L/sec}, 60 \text{ L/sec}) \] - Since fire flow is small compared to demand, the **distribution system** should be designed for **at least 16,944 L/sec**. --- # **Final Summary for Problem 2:** | Quantity | Estimated Value | |---|---| | Total water demand | 732 million L/day | | Water treatment plant capacity | ~16,944 L/sec | | Water distribution system capacity | ~16,944 L/sec | | Estimated fire flow | ~3,600 L/min (60 L/sec) | --- # **Key Assumptions:** - Fire flow factors are based on standard values. - Fire duration is 2 hours for calculation. - Peak factors are applied to demand for capacity estimation. - Fire flow and water demands are approximated proportionally. --- **End of Solution**