Certainly! Let's carefully analyze the problem step-by-step to determine the **minor head loss (h_L, minor)** in the piping system. --- # **Problem Recap:** - Water flow rate, \(Q = \text{unknown}\) (we'll find it from velocity) - Velocity of water, \(V = 5\,m/s\) - Pipe diameter, \(D = 10\,cm = .1\,m\) - Density of water, \(\rho = 998.2\,kg/m^3\) - Dynamic viscosity, \(\mu = 1.0016 \times 10^{-3}\,kg/(m \cdot s)\) - Pipe length, \(L_{total} = 30 + 60 + 12\,m = 102\,m\) - Inlet type: sharp-edged, \(K_{inlet} = .5\) - Flanged elbows: \(K_{elbow} = .3\) each - Globe valve: \(K_{valve} = 10\) (fully open) - Cross-sectional area of pipe, \(A = \frac{\pi}{4} D^2\) --- # **Step 1: Calculate the flow rate \(Q\)** Using the velocity: \[ Q = A \times V \] Where, \[ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} \times (.1)^2 \approx 7.854 \times 10^{-3}\, m^2 \] So, \[ Q = 7.854 \times 10^{-3}\, m^2 \times 5\, m/s = .03927\, m^3/s \] --- # **Step 2: Determine the Reynolds number (\(Re\)) to confirm flow regime** \[ Re = \frac{\rho V D}{\mu} \] \[ Re = \frac{998.2 \times 5 \times .1}{1.0016 \times 10^{-3}} \approx 498,200 \] Since \(Re > 400\), flow is turbulent, so Darcy-Weisbach equation applies for head loss calculations. --- # **Step 3: Calculate the head loss due to friction (Darcy-Weisbach)** \[ h_f = \frac{4fL}{D} \times \frac{V^2}{2g} \] Where \(f\) is the Darcy friction factor. For turbulent flow in smooth pipes, \(f\) can be estimated using the Colebrook equation or approximated with the Swamee-Jain equation: \[ f = \frac{.25}{\left[\log_{10} \left( \frac{\varepsilon}{3.7 D} + \frac{5.74}{Re^{.9}} \right) \right]^2} \] Given: \[ \varepsilon = .20\,mm = .0002\,m \] Calculate: \[ f = \frac{.25}{\left[\log_{10} \left( \frac{.0002}{3.7 \times .1} + \frac{5.74}{(498200)^{.9}} \right) \right]^2} \] Compute each term: \[ \frac{.0002}{3.7 \times .1} \approx .0002 / .37 \approx .00054 \] \[ \frac{5.74}{(498200)^{.9}} \approx \frac{5.74}{(498200)^{.9}} \] Estimate \(498200^{.9}\): \[ \log_{10}(498200) \approx 5.698 \] \[ .9 \times 5.698 \approx 5.128 \] \[ 10^{5.128} \approx 1.34 \times 10^{5} \] Then, \[ \frac{5.74}{1.34 \times 10^{5}} \approx 4.28 \times 10^{-5} \] Now, sum: \[ .00054 + 4.28 \times 10^{-5} \approx .0005828 \] Calculate \(\log_{10}\): \[ \log_{10}(.0005828) \approx -3.234 \] Calculate \(f\): \[ f = \frac{.25}{(-3.234)^2} = \frac{.25}{10.46} \approx .0239 \] --- # **Step 4: Calculate the total head loss (\(h_{loss}\))** Total minor head loss is the sum of losses at inlet, elbows, valve: \[ h_{minor} = h_{inlet} + 2 \times h_{elbows} + h_{valve} \] where each component is: \[ h_{component} = K \times \frac{V^2}{2g} \] --- # **Step 5: Calculate each component** - **Inlet:** \[ h_{inlet} = .5 \times \frac{V^2}{2g} \] - **Elbows (2 elbows):** \[ h_{elbows} = .3 \times \frac{V^2}{2g} \] Total for two elbows: \[ 2 \times .3 \times \frac{V^2}{2g} \] - **Valve:** \[ h_{valve} = 10 \times \frac{V^2}{2g} \] --- ### Calculate \(\frac{V^2}{2g}\): \[ g = 9.81\, m/s^2 \] \[ \frac{V^2}{2g} = \frac{(5)^2}{2 \times 9.81} = \frac{25}{19.62} \approx 1.275\, m \] --- ### Now, compute each head loss: - **Inlet:** \[ h_{inlet} = .5 \times 1.275 \approx .637\, m \] - **Elbows (2):** \[ h_{elbows} = .3 \times 1.275 \approx .383\, m \] Total for 2 elbows: \[ 2 \times .383 \approx .766\, m \] - **Valves:** \[ h_{valve} = 10 \times 1.275 \approx 12.75\, m \] --- # **Step 6: Sum all minor head losses** \[ h_{L, minor} = h_{inlet} + 2 \times h_{elbows} + h_{valve} \] \[ h_{L, minor} = .637 + .766 + 12.75 = \boxed{14.153\, m} \] --- # **Final Answer:** \[ \boxed{ \text{Total minor head loss} \approx \mathbf{14.15\,m} } \] This value represents the head loss due to all minor components (inlet, elbows, valve) in the system.