Let's break down and solve the problem step by step. --- ## **Given Data** - Pipe diameter, \( D_1 = 10 \) cm \( = .1 \) m - Flow rate, \( Q = .1 \) m³/s - Diffuser inlet diameter, \( D_2 = 10 \) cm \( = .1 \) m - Diffuser outlet diameter, \( D_3 = 20 \) cm \( = .2 \) m - Diffuser loss coefficient, \( K_L = .2 \) - Pipe is horizontal (\( z_1 = z_3 \)), plastic (\( \epsilon = \)), so smooth - Water density, \( \rho = 998 \) kg/m³ - Water viscosity, \( \mu = .001 \) kg/(m·s) - Pressure at pipe inlet = atmospheric (i.e., gauge pressure = ) at the inlet - Water discharged to atmosphere at diffuser exit Let the length of the pipe be \( L \). --- ## **Step 1: Calculate Velocities** ### **1. Inlet (Pipe and Diffuser)** \[ A_1 = A_2 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} \times (.1)^2 = .00785\, \text{m}^2 \] \[ v_1 = v_2 = \frac{Q}{A_1} = \frac{.1}{.00785} = 12.74\, \text{m/s} \] ### **2. Diffuser Outlet** \[ A_3 = \frac{\pi}{4} D_3^2 = \frac{\pi}{4} \times (.2)^2 = .0314\, \text{m}^2 \] \[ v_3 = \frac{Q}{A_3} = \frac{.1}{.0314} = 3.18\, \text{m/s} \] --- ## **Step 2: Energy Equation (Bernoulli + Head Losses)** For a horizontal pipe (\( z_1 = z_3 \)), atmospheric pressure at both ends, the energy equation between inlet (1) and outlet (3): \[ P_1 + \frac{1}{2} \rho v_1^2 = P_3 + \frac{1}{2} \rho v_3^2 + \text{Head losses} \] Since both are at atmospheric pressure (\(P_1 = P_3\)), the equation simplifies to: \[ \frac{1}{2} \rho v_1^2 = \frac{1}{2} \rho v_3^2 + \rho g \cdot h_L \] \[ \frac{v_1^2 - v_3^2}{2g} = h_L \] Where \( h_L \) is the total head loss (pipe friction + diffuser loss): \[ h_L = h_{f,\text{pipe}} + h_{L,\text{diffuser}} \] --- ### **a) Diffuser Loss** \[ h_{L,\text{diffuser}} = K_L \frac{v_2^2}{2g} \] Where \(v_2 = v_1\). ### **b) Pipe Friction Loss** \[ h_{f,\text{pipe}} = f \frac{L}{D_1} \frac{v_1^2}{2g} \] Where \( f \) is the Darcy-Weisbach friction factor (since pipe is smooth, laminar or turbulent flow will determine \( f \)). --- ## **Step 3: Reynolds Number and Friction Factor** \[ Re = \frac{\rho v_1 D_1}{\mu} \] \[ Re = \frac{998 \times 12.74 \times .1}{.001} = 1,271,932 \] So, turbulent flow. For smooth pipe, use Blasius correlation (for \( 400 < Re < 10^5 \)): \[ f = .3164 \, Re^{-.25} \] But our \( Re \) is higher; for smooth pipes at high Re, \( f \approx .018 \) (can use Moody chart). --- ## **Step 4: Substitute and Solve for \( L \)** From above: \[ \frac{v_1^2 - v_3^2}{2g} = f \frac{L}{D_1} \frac{v_1^2}{2g} + K_L \frac{v_1^2}{2g} \] Divide both sides by \( \frac{v_1^2}{2g} \): \[ \frac{v_1^2 - v_3^2}{v_1^2} = f \frac{L}{D_1} + K_L \] \[ \frac{L}{D_1} = \frac{v_1^2 - v_3^2}{v_1^2} - K_L \Bigg/ f \] But more clearly: \[ \frac{v_1^2 - v_3^2}{2g} = \left[ f \frac{L}{D_1} + K_L \right] \frac{v_1^2}{2g} \] \[ v_1^2 - v_3^2 = \left[ f \frac{L}{D_1} + K_L \right] v_1^2 \] \[ v_1^2 - v_3^2 - K_L v_1^2 = f \frac{L}{D_1} v_1^2 \] \[ v_1^2 (1 - K_L) - v_3^2 = f \frac{L}{D_1} v_1^2 \] \[ L = \frac{v_1^2 (1 - K_L) - v_3^2}{f v_1^2} D_1 \] \[ L = D_1 \left( \frac{(1 - K_L) - \frac{v_3^2}{v_1^2}}{f} \right) \] --- ## **Step 5: Plug in Values** - \( v_1 = 12.74 \) m/s - \( v_3 = 3.18 \) m/s - \( K_L = .2 \) - \( D_1 = .1 \) m - \( f \approx .018 \) Calculate \( \frac{v_3^2}{v_1^2} \): \[ v_1^2 = (12.74)^2 = 162.3 \] \[ v_3^2 = (3.18)^2 = 10.11 \] \[ \frac{v_3^2}{v_1^2} = \frac{10.11}{162.3} = .0623 \] \[ L = .1 \times \frac{(1 - .2) - .0623}{.018} \] \[ = .1 \times \frac{.8 - .0623}{.018} \] \[ = .1 \times \frac{.7377}{.018} \] \[ = .1 \times 40.98 = 4.10 \text{ m} \] --- ## **Final Answer** \[ \boxed{L = 4.1 \text{ m}} \] --- ### **Summary of Steps** 1. **Calculated velocities** at each section using flow rate and area. 2. **Applied the energy equation** (Bernoulli + head losses). 3. **Determined the friction factor** using Reynolds number and assumed smooth pipe. 4. **Set up the equation for head losses** and solved for pipe length, \( L \). **The required length of the pipe is \(\boxed{4.1 \text{ m}}\) so that the pressure at the pipe inlet is atmospheric.**