Let's break down the problem step by step: --- ### **Given Data:** - **Population (P) = 50,000** - **Average daily usage = 175 gpcd** *(gallons per capita per day)* - **Working storage = 2.5 MGD** *(million gallons)* - **Emergency storage = 3 days* - **Fire demand formula:** \( Q_{fire} = \frac{1020P}{(1-0.01P)} \) for **10 hours duration** --- ## **Step 1: Average Water Demand** **Average daily demand = Population × Per capita usage** \[ \text{Average daily demand} = 50,000 \times 175 = 8,750,000 \text{ gallons/day} \] Convert to **MGD** (million gallons per day): \[ \text{Average daily demand (MGD)} = \frac{8,750,000}{1,000,000} = 8.75 \text{ MGD} \] --- ## **Step 2: Design Flow for Water Treatment Plant** The **design flow** is typically taken as the maximum daily demand, which is usually about 1.5 to 2 times the average daily demand (unless otherwise specified). Assuming **maximum daily demand = 1.5 × average daily demand**: \[ \text{Maximum daily demand} = 1.5 \times 8.75 = 13.125 \text{ MGD} \] **Final Answer (a):** > **Design flow for water treatment plant = 13.1 MGD** (rounded to one decimal place) --- ## **Step 3: Service Tank Sizing** **Total Storage Required = Working Storage + Emergency Storage + Fire Storage** ### **A. Working Storage** Given as **2.5 MGD**. ### **B. Emergency Storage** For **3 days**: \[ \text{Emergency Storage} = 3 \times \text{Average daily demand} = 3 \times 8.75 = 26.25 \text{ MG} \] ### **C. Fire Storage** First, calculate **Q_fire**: Given: \[ Q_{fire} = \frac{1020P}{(1-0.01P)} \] Where \( P \) is in thousands. - \( P = 50 \) (since population is 50,000) So, \[ Q_{fire} = \frac{1020 \times 50}{(1-0.01 \times 50)} = \frac{51,000}{1-0.5} = \frac{51,000}{0.5} = 102,000 \text{ gallons (for 10 hours)} \] Convert to **MG (million gallons):** \[ \text{Fire Storage} = \frac{102,000}{1,000,000} = 0.102 \text{ MG} \] --- ### **Total Service Tank Storage Required:** \[ \text{Total Storage} = \text{Working} + \text{Emergency} + \text{Fire} = 2.5 + 26.25 + 0.102 = 28.852 \text{ MG} \] **Final Answer (b):** > **Service tank storage required = 28.85 MG** (rounded to two decimal places) --- ## **Step 4: Fire Demand for 10 Hours** Already calculated above: **Final Answer (c):** > **Fire demand for 10 hours = 102,000 gallons** (or 0.102 MG) --- ## **Summary of Final Answers** **a. Design flow for water treatment plant:** \[ \boxed{13.1 \text{ MGD}} \] **b. Service tank storage required:** \[ \boxed{28.85 \text{ MG}} \] **c. Fire demand for 10 hours:** \[ \boxed{102,\!000 \text{ gallons} \ (\text{or } 0.102 \text{ MG})} \] --- **Let me know if you need further clarifications!**

Let's carefully approach this problem step by step, incorporating the new fire demand formula: --- ### **Given Data:** - Population, \( P = 50,000 \) - Per capita daily usage, \( 175\, \text{gpcd} \) - Working storage, \( 2.5\, \text{MG} \) - Emergency storage, 3 days' demand - Fire demand formula: \[ Q_{fire} = 1020 \times \sqrt{P} \times (1 - 0.01 \times \sqrt{P}) \] *Note:* The formula is applied for the fire demand over 10 hours. --- ## **Step 1: Calculate the average daily demand** \[ \text{Daily demand} = P \times \text{per capita usage} = 50,000 \times 175 = 8,750,000\, \text{gallons} \] Convert to MGD: \[ \text{Average daily demand} = \frac{8,750,000}{1,000,000} = 8.75\, \text{MGD} \] --- ## **Step 2: Determine the maximum daily flow for the treatment plant** Assuming a peak factor of 1.5: \[ \text{Design flow} = 1.5 \times 8.75 = 13.125\, \text{MGD} \] *Final answer:* ### **a. Design flow for water treatment plant = \(\boxed{13.1\, \text{MGD}}\)** --- ## **Step 3: Calculate fire demand \(Q_{fire}\)** Calculate \(\sqrt{P}\): \[ \sqrt{50,000} = \sqrt{5 \times 10^4} \approx 223.607 \] Calculate \(Q_{fire}\): \[ Q_{fire} = 1020 \times 223.607 \times (1 - 0.01 \times 223.607) \] Calculate \(0.01 \times 223.607\): \[ 0.01 \times 223.607 \approx 2.236 \] So, \[ Q_{fire} = 1020 \times 223.607 \times (1 - 2.236) = 1020 \times 223.607 \times (-1.236) \] Notice that: \[ 1 - 2.236 = -1.236 \] which yields a **negative** fire demand, which is physically nonsensical. This indicates that at high population sizes, the formula produces a negative value, suggesting it's valid only for certain ranges of \( P \). **In practice:** - For \( P \) such that \( 0.01 \times \sqrt{P} < 1 \), the formula is positive. - For \( P = 50,000 \), \( 0.01 \times 223.607 \approx 2.236 > 1 \), leading to negative demand. **Therefore, the formula is only valid for \( P \) where:** \[ 0.01 \times \sqrt{P} < 1 \implies \sqrt{P} < 100 \implies P < 10,000 \] Since \( P = 50,000 \) exceeds this, the formula isn't valid for this population. --- ## **Conclusion:** - The fire demand formula **is not applicable** for \( P = 50,000 \) because it yields a negative value. - Typically, for large populations exceeding the formula's valid range, **a standard fire flow of 3,000 to 4,000 GPM** (or approximately 4.3 to 5.8 MGD) is used. --- ## **Final Step: Fire demand estimate** **Using standard practice**: Assuming fire demand for 10 hours: \[ \text{Fire demand} \approx 4\, \text{MGD} \] *(This is a typical fire flow rate for a community of this size)* --- ## **Step 4: Service tank storage** Total storage: \[ \text{Total} = \text{Working storage} + (\text{Emergency storage}) + (\text{Fire storage}) \] - **Working storage:** 2.5 MG - **Emergency storage:** \( 3 \times 8.75 = 26.25\, \text{MG} \) - **Fire storage:** 4 MG (assumed) **Total:** \[ 2.5 + 26.25 + 4 = \boxed{32.75\, \text{MG}} \] --- ## **Final answers:** ### **a. Design flow for water treatment plant:** \[ \boxed{13.1\, \text{MGD}} \] ### **b. Service tank storage:** \[ \boxed{32.75\, \text{MG}} \] ### **c. Fire demand over 10 hours:** \[ \boxed{4\, \text{MG}} \] --- **Note:** Since the fire demand formula yields invalid results at high populations, using standard community fire flow rates is advisable.

Let's carefully analyze the problem step by step, incorporating the formula: \[ Q_{\text{fire}} = 1020 \times \sqrt{P} \times (1 - 0.01 \times \sqrt{P}) \] where \( P \) is population in thousands. --- ### **Given Data:** - Population: \( P = 50,000 \) - Per capita daily usage: 175 gpcd - Working storage: 2.5 MG - Emergency storage: 3 days' demand - Fire demand is calculated for a 10-hour duration. --- ### **Step 1: Convert population to thousands** \[ P_{thousands} = \frac{50,000}{1,000} = 50 \] --- ### **Step 2: Calculate \(\sqrt{P}\)** \[ \sqrt{50} \approx 7.07 \] --- ### **Step 3: Calculate fire demand \(Q_{\text{fire}}\)** \[ Q_{\text{fire}} = 1020 \times 7.07 \times (1 - 0.01 \times 7.07) \] Calculate \(0.01 \times 7.07\): \[ 0.01 \times 7.07 = 0.0707 \] Now, \[ Q_{\text{fire}} = 1020 \times 7.07 \times (1 - 0.0707) = 1020 \times 7.07 \times 0.9293 \] Compute step-by-step: 1. \(1020 \times 7.07 \approx 7217.4\) 2. Multiply by 0.9293: \[ 7217.4 \times 0.9293 \approx 6714.4 \text{ gallons} \] This is the fire demand for **10 hours**. --- ### **Step 4: Convert fire demand to MG** \[ Q_{\text{fire}} = \frac{6714.4}{1,000,000} \approx 0.006714 \text{ MG} \] or approximately **0.0067 MG**. --- ### **Step 5: Calculate total storage requirement** 1. **Average daily demand:** \[ \text{Daily demand} = 50,000 \times 175 = 8,750,000 \text{ gallons} \] Convert to MG: \[ \frac{8,750,000}{1,000,000} = 8.75 \text{ MG} \] 2. **Emergency storage:** \[ 3 \times 8.75 = 26.25 \text{ MG} \] 3. **Total storage needed:** \[ \text{Working storage} + \text{Emergency storage} + \text{Fire storage} = 2.5 + 26.25 + 0.0067 \approx \boxed{28.76 \text{ MG}} \] --- ### **Final Answers:** **a. Water treatment plant design flow:** Maximum daily demand (assuming 1.5 times average): \[ 1.5 \times 8.75 = \boxed{13.1\, \text{MGD}} \] **b. Service tank storage:** \[ \boxed{28.76\, \text{MG}} \] **c. Fire demand for 10 hours:** \[ \boxed{0.0067\, \text{MG}} \quad \text{or} \quad 6,714 \text{ gallons} \] --- **Summary:** | Parameter | Result | |------------|---------| | **Design flow** | **13.1 MGD** | | **Service tank storage** | **28.76 MG** | | **Fire demand (10 hours)** | **6,714 gallons (0.0067 MG)** | Feel free to ask for further clarifications!

Let's carefully analyze and solve the problem step by step. --- ### **Given Data:** - **Road type:** Two-lane undivided rural road - **Lane width:** 3.5 m per lane - **Shoulder width:** 1.5 m - **Total width of each carriageway (including shoulders):** \( \text{Lane} + \text{Shoulder} = 3.5 + 1.5 = 5.0\, \text{m} \) - **Bearings of tangents:** - First tangent: 45° - Second tangent: 105° - **Design speed:** 80 km/hr - **Maximum E (superelevation factor):** Use maximum E for the speed/type of road (typical maximum E for rural roads: 0.12 to 0.15; we'll assume **E = 0.15**) - **Maximum desirable f (friction factor):** Approximately 0.30 (standard for design) - **Large trees are located 1.25 m from the inside edge of the road at mid-curve.** --- ### **Step 1: Find the deflection angle Δ between the tangents** The bearings: - \( \text{Bearing}_1 = 45^\circ \) - \( \text{Bearing}_2 = 105^\circ \) **Difference in bearings:** \[ \Delta = |\text{Bearing}_2 - \text{Bearing}_1| = |105^\circ - 45^\circ| = 60^\circ \] --- ### **Step 2: Calculate the length of the curve (L)** The **length of the curve (L)** is related to the radius (R) and the central angle (Δ): \[ L = R \times \theta \] where \( \theta \) is in **radians**: \[ \theta = \frac{\Delta \times \pi}{180} = \frac{60^\circ \times \pi}{180} = \frac{\pi}{3} \approx 1.047\, \text{radians} \] --- ### **Step 3: Find the minimum radius R for the design speed** The **minimum radius R** for a curve, considering maximum E and f, is given by the standard formula: \[ R = \frac{V^2}{g \times (E + f)} \] where: - \( V \) = design speed in m/sec - \( g \) = acceleration due to gravity ≈ 9.81 m/sec² - \( E \) = maximum superelevation (assumed 0.15) - \( f \) = maximum friction factor (assumed 0.30) First, convert speed: \[ V = 80\, \text{km/hr} = \frac{80,000}{3600} \approx 22.22\, \text{m/sec} \] Calculate \( R \): \[ R = \frac{(22.22)^2}{9.81 \times (0.15 + 0.30)} = \frac{493.8}{9.81 \times 0.45} = \frac{493.8}{4.410} \approx 112\, \text{meters} \] **Rounded up to nearest meter:** \[ \boxed{112\, \text{meters}} \] --- ### **Step 4: Calculate the length of the curve (L)** \[ L = R \times \theta = 112 \times 1.047 \approx 117.2\, \text{meters} \] --- ### **Step 5: Determine the midpoint of the curve** - **Arc length at mid-curve:** Half of total length: \[ L/2 = 117.2/2 \approx 58.6\, \text{meters} \] - **Radius of the curve (R):** 112 m - **Distance from the inside edge to the mid-curve point:** The **distance from the inside edge of the curve to the mid-arc point** is: \[ \text{Offset} = R - \text{inside edge to curve center} \] But more practically, the **inside edge of the road is at a radius R**, and trees are located 1.25 m from the inside edge, i.e., at: \[ \text{Tree location from centerline} = R - 1.25\, \text{m} \] --- ### **Step 6: Calculate how much to cut back the trees** - The **trees are located 1.25 m from the inside edge**. - The **mid-arc point** is at a radius: \[ R_{\text{trees}} = R - 1.25 = 112 - 1.25 = 110.75\, \text{meters} \] - The **distance from the center of the curve to the trees** at mid-curve is approximately **R - 1.25 m**. - To clear the trees, the **inside edge of the road** must be moved inward at the mid-curve point. - The **additional cutback** needed is the **difference between the existing radius (R)** and the **radius to the trees (R - 1.25 m)**. Since the **trees are 1.25 m from the inside edge**, the **cutback** necessary is **1.25 meters**. --- ### **Final answer:** **The trees need to be cut back by approximately \(\boxed{1.25\, \text{meters}}\).** --- ## **Summary:** | Parameter | Value | |------------|---------| | Minimum radius R | **112 meters** | | Mid-curve arc length | **117 meters** | | Trees need to be cut back | **1.25 meters** | --- **Note:** The key step is recognizing the radius and the position of trees relative to the inside edge; the cutback equals the distance from the inside edge to the trees, which is 1.25 m.