# Pipe Flow Problem with Updated Pressure at A ## Given Data - **Diameter of pipe, D \)** 50 mm = . m - **Length between pressure gauges, \( L \)**: 6 m - **Temperature, \( T \)**: 20°C - **Pressure at A, P_A \)**: **17 kPa** (updated) = 17,000 Pa - **Pressure at B, \( P_B \)**: 10.5 kPa = 10,500 Pa - **Dynamic viscosity, \( \mu \)**: \( 1.9 \times 10^{-3} \) Pa·s - **Density, \( \rho \)**: 814 kg/m³ - **Gravity, \( g \)**: 9.81 m/s² --- ## Step 1: Head Loss Calculation ### Using Bernoulli's Equation (Horizontal Pipe, Constant Velocity): \[ \frac{P_A}{\rho g} + \frac{V^2}{2g} = \frac{P_B}{\rho g} + \frac{V^2}{2g} + h_L \] Since elevation and velocity are constant: \[ h_L = \frac{P_A - P_B}{\rho g} \] Plug in the values: \[ h_L = \frac{17,000 - 10,500}{814 \times 9.81} \] \[ h_L = \frac{6,500}{7,987.34} = .814 \text{ m} \] --- ## Step 2: Assume Laminar Flow (for Checking) \[ h_L = \frac{32 \mu V L}{\rho g D^2} \] Plug in values: \[ .814 = \frac{32 \times 1.9 \times 10^{-3} \times V \times 6}{814 \times 9.81 \times (.05)^2} \] Calculate the denominator: \[ 814 \times 9.81 \times .0025 = 19.96965 \] Numerator: \[ 32 \times 1.9 \times 10^{-3} \times 6 = .3648 \] So: \[ .814 = \frac{.3648 V}{19.96965} \] \[ .814 \times 19.96965 = .3648 V \] \[ 16.25 = .3648 V \] \[ V = \frac{16.25}{.3648} \approx 44.55\, \text{m/s} \] ### Reynolds Number (Laminar Check) \[ Re = \frac{\rho V D}{\mu} \] \[ Re = \frac{814 \times 44.55 \times .05}{1.9 \times 10^{-3}} \] \[ 814 \times 44.55 \times .05 = 1,812.4 \] \[ Re = \frac{1,812.4}{.0019} \approx 954947 \] **Since \( Re \gg 200 \), flow is not laminar.** --- ## Step 3: Turbulent Flow Head Loss (Darcy-Weisbach Equation) \[ h_L = f \frac{L}{D} \frac{V^2}{2g} \] Where friction factor for smooth pipes (Blasius correlation for turbulent flow): \[ f = .316 \times Re^{-.25} \] Let's solve iteratively: ### Express \( h_L \) in terms of \( V \): \[ .814 = f \frac{6}{.05} \frac{V^2}{2 \times 9.81} \] \[ .814 = f \times 120 \times \frac{V^2}{19.62} \] \[ .814 = f \times 6.116 \times V^2 \] \[ V^2 = \frac{.814}{f \times 6.116} \] \[ V = \sqrt{\frac{.814}{f \times 6.116}} \] But \( f \) depends on \( V \) via \( Re \): \[ Re = \frac{814 \times V \times .05}{1.9 \times 10^{-3}} = 21,421 \times V \] \[ f = .316 \times (21,421 \times V)^{-.25} \] Set up the equation for \( V \): \[ V = \sqrt{ \frac{.814}{.316 \times (21,421 V)^{-.25} \times 6.116} } \] Let's solve for \( V \) numerically: --- ### Iterative Solution 1. **Initial guess**: \( V = 2 \) m/s - \( Re = 21,421 \times 2 = 42,842 \) - \( f = .316 \times (42,842)^{-.25} \) - \( \log_{10}(42,842) = 4.632 \) - \( (42,842)^{-.25} = 10^{-.25 \times 4.632} = 10^{-1.158} = .0695 \) - \( f = .316 \times .0695 = .02195 \) - \( V = \sqrt{ \frac{.814}{.02195 \times 6.116} } = \sqrt{ \frac{.814}{.1343} } = \sqrt{6.06} = 2.46 \) m/s 2. **Second iteration**: \( V = 2.46 \) m/s - \( Re = 21,421 \times 2.46 = 52,693 \) - \( \log_{10}(52,693) = 4.72 \) - \( (52,693)^{-.25} = 10^{-.25 \times 4.72} = 10^{-1.18} = .066 \) - \( f = .316 \times .066 = .02086 \) - \( V = \sqrt{ \frac{.814}{.02086 \times 6.116} } = \sqrt{ \frac{.814}{.1276} } = \sqrt{6.38} = 2.52 \) m/s 3. **Third iteration**: \( V = 2.52 \) m/s - \( Re = 21,421 \times 2.52 = 53,985 \) - \( \log_{10}(53,985) = 4.732 \) - \( (53,985)^{-.25} = 10^{-.25 \times 4.732} = 10^{-1.183} = .0656 \) - \( f = .316 \times .0656 = .02074 \) - \( V = \sqrt{ \frac{.814}{.02074 \times 6.116} } = \sqrt{ \frac{.814}{.127} } = \sqrt{6.41} = 2.53 \) m/s **Value is converging.** --- ## Step 4: Final Reynolds Number \[ Re = \frac{814 \times 2.53 \times .05}{1.9 \times 10^{-3}} \] \[ 814 \times 2.53 \times .05 = 102.855 \] \[ Re = \frac{102.855}{.0019} \approx 54,134 \] --- ## **Final Answers** **Reynolds number:** \[ \boxed{Re = 54,\!134} \] **Flow is turbulent.** --- ## **Summary Table** | Quantity | Value | |-----------------------|----------------| | \( h_L \) | .814 m | | \( V \) (turbulent) | 2.53 m/s | | \( Re \) | 54,134 | | Flow Regime | Turbulent |