Let's break down the problem step by step: ## **Given Data** - Specific gravity of oil, \( S = .92 \) - Viscosity, \( \mu = .00065 \) lb·sec/ft² - Flow rate, \( Q = 61.04 \) cfs - Pipe diameter, \( D = 12 \) in \( = 1 \) ft - Pipe roughness, \( \epsilon = .00013 \) ft --- ## **Step 1: Find the Density of Oil** Specific gravity \( S = \frac{\rho}{\rho_{water}} \) - Density of water, \( \rho_{water} = 62.4 \) lb/ft³ \[ \rho = S \times \rho_{water} = .92 \times 62.4 = 57.408 \text{ lb/ft}^3 \] --- ## **Step 2: Find the Velocity** Area of pipe, \( A \): \[ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (1)^2 = .7854 \text{ ft}^2 \] Velocity, \( V \): \[ V = \frac{Q}{A} = \frac{61.04}{.7854} = 77.76 \text{ ft/s} \] --- ## **Step 3: Calculate Reynolds Number** \[ Re = \frac{\rho V D}{\mu} \] Plug in the values: \[ Re = \frac{57.408 \times 77.76 \times 1}{.00065} \] \[ Re = \frac{4465.3}{.00065} = 6,870,385 \] --- ## **Step 4: Find Relative Roughness** \[ \frac{\epsilon}{D} = \frac{.00013}{1} = .00013 \] --- ## **Step 5: Calculate Friction Factor (using Colebrook-White equation or Moody chart)** For \( Re \gg 400 \), it's turbulent. Use **Swamee-Jain equation** (approximation): \[ f = .25 \left[ \log_{10} \left( \frac{\epsilon}{3.7D} + \frac{5.74}{Re^{.9}} \right) \right]^{-2} \] \[ \frac{\epsilon}{3.7D} = \frac{.00013}{3.7 \times 1} = .0000351 \] \[ \frac{5.74}{Re^{.9}} = \frac{5.74}{(6,870,385)^{.9}} \approx \frac{5.74}{1,340,000} = .00000428 \] \[ f = .25 \left[ \log_{10}(.0000351 + .00000428) \right]^{-2} \] \[ f = .25 \left[ \log_{10}(.00003938) \right]^{-2} \] \[ \log_{10}(.00003938) \approx -4.405 \] \[ f = .25 \left[ -4.405 \right]^{-2} \] \[ f = .25 \times (.0515) \] \[ f \approx .0129 \] --- ## **Step 6: Friction Head Loss (\( h_f \))** Darcy-Weisbach equation: \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] Assume \( L = 1 \) ft for head loss per foot length: \[ h_f = .0129 \times \frac{1}{1} \times \frac{(77.76)^2}{2 \times 32.174} \] \[ h_f = .0129 \times \frac{605}{64.348} \] \[ = .0129 \times 94.036 \] \[ = 1.21 \text{ ft per ft} \] --- ## **Step 7: Shear Stress at the Pipe Wall (\( \tau_w \))** \[ \tau_w = f \frac{\rho V^2}{8} \] \[ \tau_w = .0129 \frac{57.408 \times (77.76)^2}{8} \] \[ = .0129 \frac{57.408 \times 605}{8} \] \[ = .0129 \frac{347,318}{8} \] \[ = .0129 \times 43,415 \] \[ = 560 \text{ lb/ft}^2 \] --- ## **Step 8: Pipe Behavior (Rough, Transitional, Smooth?)** - Since \( Re \gg 400 \) (turbulent) - Relative roughness is small, but \( Re \) is extremely high For such high \( Re \), friction factor is dominated by roughness (\( f \) independent of \( Re \)), so **fully rough** regime. --- ## **Final Answers** **1. Friction head loss**: \[ h_f \approx 1.21 \text{ ft per ft length of pipe} \] **2. Shear stress at the wall**: \[ \tau_w \approx 560 \text{ lb/ft}^2 \] **3. Pipe behavior:** **Fully rough** regime --- ### **Summary Table** | Quantity | Value | |-----------------------|--------------------------| | Reynolds number, \( Re \) | \( 6,870,385 \) | | Friction factor, \( f \) | \( .0129 \) | | Friction head loss, \( h_f \) | \( 1.21 \) ft/ft | | Shear stress, \( \tau_w \) | \( 560 \) lb/ft² | | Pipe regime | Fully Rough |