Let's solve **Problem 6** step by step for each method: --- ## **Given Data** - **Q** = 150 cfs (cubic feet per second) - **D** = 72 in. = 6 ft (diameter) - **L** = 10,000 ft (length) - **Pipe material**: Steel (assume commercial steel, typical roughness, \( \epsilon \approx .00015 \) ft) - **Water at 60°F**: \( \nu \approx 1.217 \times 10^{-5} \) ft²/s, \( g = 32.2 \) ft/s² --- ## **Step 1: Convert Units and Calculate Area, Velocity** - **Pipe area**: \( A = \frac{\pi}{4}D^2 = \frac{\pi}{4} (6)^2 = 28.274 \) ft² - **Velocity**: \( V = \frac{Q}{A} = \frac{150}{28.274} = 5.31 \) ft/s --- ## **a. Darcy-Weisbach Equation** ### **Step 1: Calculate Reynolds Number** \[ Re = \frac{VD}{\nu} = \frac{5.31 \times 6}{1.217 \times 10^{-5}} = \frac{31.86}{1.217 \times 10^{-5}} \approx 2,619,000 \] (Laminar if \( Re < 200 \), turbulent otherwise; this is turbulent.) ### **Step 2: Estimate Friction Factor \( f \)** Use Colebrook-White or Moody diagram for commercial steel: \[ \frac{\epsilon}{D} = \frac{.00015}{6} = 2.5 \times 10^{-5} \] For high Re, steel pipe, friction factor \( f \approx .012 \) (from Moody diagram). ### **Step 3: Darcy-Weisbach Head Loss Formula** \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] Plug in values: \[ h_f = .012 \cdot \frac{10,000}{6} \cdot \frac{(5.31)^2}{2 \cdot 32.2} \] \[ \frac{10,000}{6} = 1666.67 \] \[ (5.31)^2 = 28.21 \] \[ 2 \cdot 32.2 = 64.4 \] \[ \frac{28.21}{64.4} = .438 \] \[ h_f = .012 \times 1666.67 \times .438 = 8.75 \, \text{ft} \] --- ## **b. Hazen-Williams Formula** \[ h_f = 10.67 \frac{L}{C^{1.85} D^{4.87}} Q^{1.85} \] Where: - \( L \) in ft - \( D \) in ft - \( Q \) in cfs - \( C \) for steel pipe: typically \( C = 120 \) \[ h_f = 10.67 \cdot \frac{10,000}{120^{1.85} \cdot 6^{4.87}} \cdot 150^{1.85} \] Calculate each term: - \( 120^{1.85} \approx 5444 \) - \( 6^{4.87} \approx 632 \) - \( 150^{1.85} \approx 6836 \) So, \[ h_f = 10.67 \cdot \frac{10,000}{5444 \times 632} \cdot 6836 \] \[ = 10.67 \cdot \frac{10,000}{34,409,080} \cdot 6836 \] \[ = 10.67 \cdot .0002905 \cdot 6836 \] \[ = 10.67 \cdot 1.987 \] \[ = 21.2 \, \text{ft} \] --- ## **c. Swamee-Jain Formula** The Swamee-Jain formula provides an explicit equation for the Darcy friction factor: \[ f = .25 \left[ \log_{10} \left( \frac{\epsilon}{3.7D} + \frac{5.74}{Re^{.9}} \right) \right]^{-2} \] Plug in values: - \( \epsilon = .00015 \) ft - \( D = 6 \) ft - \( Re = 2,619,000 \) \[ \frac{\epsilon}{3.7D} = \frac{.00015}{3.7 \times 6} = 6.76 \times 10^{-6} \] \[ \frac{5.74}{Re^{.9}} = \frac{5.74}{(2.619 \times 10^6)^{.9}} \approx \frac{5.74}{559,000} = 1.027 \times 10^{-5} \] \[ \text{Sum:} \approx 1.70 \times 10^{-5} \] \[ \log_{10}(1.70 \times 10^{-5}) = -4.77 \] \[ f = .25 [ -4.77 ]^{-2} = .25 / (22.76) = .011 \] Now, use Darcy-Weisbach formula from part (a) with \( f = .011 \): \[ h_f = .011 \times 1666.67 \times .438 = 8.06 \, \text{ft} \] --- ## **Summary Table** | Method | Head Loss (\( h_f \), ft) | |-------------------------------|--------------------------| | Darcy-Weisbach (Moody) | 8.75 | | Hazen-Williams | 21.2 | | Swamee-Jain | 8.06 | --- # **Final Answers** 1. **Darcy-Weisbach (Moody):** \( h_f \approx 8.75 \) ft 2. **Hazen-Williams:** \( h_f \approx 21.2 \) ft 3. **Swamee-Jain:** \( h_f \approx 8.06 \) ft *The Hazen-Williams formula typically gives higher head losses for large, smooth pipes compared to Darcy-Weisbach-based methods.* --- ### If you need a solution for problem 7, let me know!