Let's solve the problem step by step! --- ## **Given Data** - Pipe length, \( L = 100 \) m - Elevation difference, \( h = 200 \) m - Friction factor, \( f = .02 \) - Thickness, \( t_w = 20 \) mm \( = .02 \) m - Diameter, \( D = 2. \) m - For water: - Density, \( \rho = 100 \) kg/m³ - Specific weight, \( \gamma = 9806 \) N/m³ - Bulk modulus, \( E = 2.2 \times 10^9 \) Pa - For pipe: - Bulk modulus, \( E_p = 1.9 \times 10^{11} \) Pa - Max allowable stress: \( \tau = 1.3 \times 10^8 \) Pa - Gravity, \( g = 9.81 \) m/s² --- ## **(a) Flow Rate and Steady State Velocity** ### **1. Energy Equation (Bernoulli with head loss):** \[ z_1 + \frac{P_1}{\gamma} + \frac{V_1^2}{2g} = z_2 + \frac{P_2}{\gamma} + \frac{V_2^2}{2g} + h_f \] - Reservoir: \( z_1 = 200 \) m, \( P_1 = \), \( V_1 \approx \) - Outlet: \( z_2 = \), \( P_2 = \), \( V_2 = V \) - Head loss due to friction: \( h_f = f \frac{L}{D} \frac{V^2}{2g} \) So, \[ 200 = \frac{V^2}{2g} + + f \frac{L}{D} \frac{V^2}{2g} \] \[ 200 = \frac{V^2}{2g} \left[1 + f \frac{L}{D}\right] \] \[ V^2 = 2g \cdot 200 / \left[1 + f \frac{L}{D}\right] \] \[ V^2 = 2 \cdot 9.81 \cdot 200 / (1 + .02 \cdot \frac{100}{2}) \] \[ = 3924 / (1 + 10) \] \[ = 3924 / 11 = 356.73 \] \[ V = \sqrt{356.73} \approx 18.9\, \text{m/s} \] **Flow Rate:** \[ Q = A V = \frac{\pi}{4} D^2 V = \frac{\pi}{4} \cdot (2)^2 \cdot 18.9 \approx 3.142 \cdot 18.9 \approx 59.4\, \text{m}^3/\text{s} \] --- ## **(b) Celerity \( c \) and Critical Time of Closure** ### **1. Celerity \( c \):** \[ c = \sqrt{ \frac{K}{\rho \left[1 + \frac{K D}{E_p t_w}\right]} } \] Where: - \( K \) = bulk modulus of water = \( 2.2 \times 10^9 \) Pa - \( E_p \) = pipe modulus = \( 1.9 \times 10^{11} \) Pa \[ c = \sqrt{ \frac{2.2 \times 10^9}{100 \left[1 + \frac{2.2 \times 10^9 \cdot 2}{1.9 \times 10^{11} \cdot .02}\right]} } \] Calculate denominator inside the brackets: \[ \frac{K D}{E_p t_w} = \frac{2.2 \times 10^9 \cdot 2}{1.9 \times 10^{11} \cdot .02} = \frac{4.4 \times 10^9}{3.8 \times 10^9} = 1.158 \] So, \[ c = \sqrt{ \frac{2.2 \times 10^9}{100 (1 + 1.158)} } = \sqrt{ \frac{2.2 \times 10^9}{100 \times 2.158} } = \sqrt{ \frac{2.2 \times 10^9}{2158} } = \sqrt{1.019 \times 10^6} \approx 101\, \text{m/s} \] ### **2. Critical Time of Closure \( t_c \):** \[ t_c = \frac{2L}{c} = \frac{2 \times 100}{101} \approx 1.98\, \text{seconds} \] --- ## **(c) Pressure Increase and Total Pressure at Valve (Valve closed in .8s < \( t_c \))** When closure time \( t_{close} < t_c \), maximum surge: \[ \Delta P = \rho c V \] \[ \Delta P = 100 \cdot 101 \cdot 18.9 = 19,089,000 \approx 1.91 \times 10^7\, \text{Pa} \] **Total Pressure at Valve:** The static pressure at the valve is due to elevation: \[ P_{static} = \rho g h = 100 \cdot 9.81 \cdot 200 = 1,962,000\, \text{Pa} \] \[ P_{total} = P_{static} + \Delta P = 1,962,000 + 19,089,000 = 21,051,000\, \text{Pa} \] --- ## **(d) Pressure Surge and Total Pressure (Valve closes in 10s > \( t_c \))** For \( t_{close} > t_c \): \[ \Delta P = \frac{2 \rho L V}{t_{close}} \] \[ \Delta P = \frac{2 \cdot 100 \cdot 100 \cdot 18.9}{10} = \frac{37,800,000}{10} = 3,780,000\, \text{Pa} \] \[ P_{total} = P_{static} + \Delta P = 1,962,000 + 3,780,000 = 5,742,000\, \text{Pa} \] --- ## **(e) Wall Stress at \( t = 10 \) s and Pipe Safety** \[ \tau = \frac{P D}{2 t_w} \] Where \( P = P_{total} \) from part (d): \[ \tau = \frac{5,742,000 \cdot 2}{2 \cdot .02} = \frac{11,484,000}{.04} = 287,100,000\, \text{Pa} = 2.87 \times 10^8\, \text{Pa} \] **Maximum allowable stress:** \( 1.3 \times 10^8 \) Pa **Conclusion:** \( \tau = 2.87 \times 10^8 \) Pa **exceeds** the maximum allowed tension stress. **The pipe cannot withstand the surge pressure.** --- ## **Summary Table** | Part | Quantity | Value | |------|-------------------------------------|--------------------------| | (a) | Steady velocity, \( V \) | 18.9 m/s | | | Flow rate, \( Q \) | 59.4 m³/s | | (b) | Celerity, \( c \) | 101 m/s | | | Critical closure time, \( t_c \) | 1.98 s | | (c) | Pressure surge (.8s closure) | \( 1.91 \times 10^7 \) Pa| | | Total pressure at valve | \( 2.11 \times 10^7 \) Pa| | (d) | Pressure surge (10s closure) | \( 3.78 \times 10^6 \) Pa| | | Total pressure at valve | \( 5.74 \times 10^6 \) Pa| | (e) | Pipe wall stress (\( t=10 \) s) | \( 2.87 \times 10^8 \) Pa| | | Max allowable stress | \( 1.3 \times 10^8 \) Pa | | | **Safe?** | **NO** | --- Let me know if you need detailed derivations for any individual step!