Let's solve the problem **step by step**. --- ## **Given Data** - **Pipe diameter, \( D \) = 4 cm = .04 m** - **Pipe length, \( L \) = 50 m** - **Flow rate, \( Q \) = .006 m³/s** - **Pressure at B, \( P_B \) = 192 kPa = 192,000 Pa** - **Horizontal pipe** (difference in elevation is zero) - **Fittings:** 1 fully open gate valve, 2 elbows (\( R/D = 4 \)) - **Wrought iron pipe** - **Water at room temperature** (Assume density \( \rho = 100 \) kg/m³, viscosity \( \mu = 1 \times 10^{-3} \) Pa.s) --- ## **Step 1: Find Velocity** \[ Q = V \cdot A \implies V = \frac{Q}{A} \] \[ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (.04)^2 = .001257 \text{ m}^2 \] \[ V = \frac{.006}{.001257} = 4.775 \text{ m/s} \] --- ## **Step 2: Calculate Reynolds Number** \[ Re = \frac{\rho V D}{\mu} = \frac{100 \times 4.775 \times .04}{.001} = 191,000 \] This is turbulent flow. --- ## **Step 3: Find Head Losses** ### **a) Major Loss (Darcy-Weisbach):** \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] #### **i. Find f (Darcy friction factor):** Assume **wrought iron pipe, commercial steel**. For turbulent flow, estimate \( f \) using the Swamee-Jain equation: \[ f = .25 \left[ \log_{10} \left( \frac{\epsilon}{3.7D} + \frac{5.74}{Re^{.9}} \right) \right]^{-2} \] Wrought iron roughness, \( \epsilon \approx .00015 \) m. \[ \frac{\epsilon}{3.7D} = \frac{.00015}{3.7 \times .04} = .00101 \] \[ \frac{5.74}{Re^{.9}} = \frac{5.74}{(191,000)^{.9}} \approx \frac{5.74}{39,300} = .000146 \] \[ \log_{10}(.00101 + .000146) = \log_{10}(.001156) = -2.937 \] \[ f = .25 \left(-2.937\right)^{-2} = .25 \times (.116) = .029 \] --- ### **b) Minor Losses (\( K \)):** \[ h_{minor} = K \frac{V^2}{2g} \] - **Gate valve (fully open):** \( K_{\text{valve}} \approx .15 \) - **Elbow (R/D=4):** \( K_{\text{elbow}} \approx .21 \) (from tables) - **Total K:** \( K_{total} = .15 + 2 \times .21 = .57 \) --- ### **c) Total Head Loss** \[ h_{total} = h_f + h_{minor} \] Calculate each: \[ h_f = f \frac{L}{D} \frac{V^2}{2g} = .029 \times \frac{50}{.04} \times \frac{(4.775)^2}{2 \times 9.81} \] \[ \frac{50}{.04} = 125 \] \[ (4.775)^2 = 22.81 \] \[ 2g = 19.62 \] \[ h_f = .029 \times 125 \times \frac{22.81}{19.62} \] \[ = .029 \times 125 \times 1.163 \] \[ = .029 \times 1453.75 = 42.16 \] \[ h_{minor} = .57 \times \frac{22.81}{19.62} = .57 \times 1.163 = .663 \] \[ h_{total} = 42.16 + .663 = 42.82 \text{ m} \] --- ## **Step 4: Apply Energy Equation** (Bernoulli, horizontal pipe, neglecting velocity change) \[ \frac{P_A}{\rho g} = \frac{P_B}{\rho g} + h_{total} \] \[ P_A = P_B + \rho g h_{total} \] \[ P_A = 192,000 + 100 \times 9.81 \times 42.82 \] \[ = 192,000 + 419,988 = 611,988 \text{ Pa} \] \[ = 612 \text{ kPa} \quad (\text{rounded to 3 significant digits}) \] --- ## **## Final Answer** \[ \boxed{P_A = 612\, \text{kPa}} \] --- ### **Summary of Steps** 1. **Find velocity from flow rate and area.** 2. **Calculate Reynolds number (turbulent flow).** 3. **Estimate friction factor using Swamee-Jain.** 4. **Calculate major and minor losses.** 5. **Apply energy equation to solve for \( P_A \).**