Let's solve the problem step-by-step. --- ## **Given Data** - **Reservoir height above pipeline:** \( H = 1.5 \) m - **Pipeline sections:** - Section 1: \( D_1 = .24 \) m, \( L_1 = 3 \) m - Section 2: \( D_2 = .1 \) m, \( L_2 = 1 \) m - Section 3: \( D_3 = .12 \) m, \( L_3 = 2 \) m - **Water temperature:** \( T = 10^\circ C \) (for part b) - **Assume:** Outlet discharges to atmospheric pressure. --- # **a) Neglecting Losses (Ideal Fluid)** ### **Step 1: Apply Bernoulli's Equation** Between the water surface (point ) and the outlet (point 3): \[ P_ + \frac{1}{2}\rho v_^2 + \rho g z_ = P_3 + \frac{1}{2}\rho v_3^2 + \rho g z_3 \] - \(P_ = P_3\) (both open to atmosphere) - \(v_ \approx \) (large reservoir) - \(z_ - z_3 = H\) \[ \rho g H = \frac{1}{2}\rho v_3^2 \] \[ g H = \frac{1}{2}v_3^2 \] \[ v_3 = \sqrt{2gH} \] ### **Step 2: Calculate Discharge** \[ v_3 = \sqrt{2 \cdot 9.81 \cdot 1.5} = \sqrt{29.43} = 5.425\, \text{m/s} \] \[ Q = v_3 \cdot A_3 \] \[ A_3 = \frac{\pi}{4} D_3^2 = \frac{\pi}{4} (.12)^2 = .0113\, \text{m}^2 \] \[ Q = 5.425 \cdot .0113 = .0613\, \text{m}^3/\text{s} \] --- **Final answers for part (a):** \[ \boxed{v_3 = 5.43\, \text{m/s}} \] \[ \boxed{Q = .0613\, \text{m}^3/\text{s}} \] --- # **b) Considering Losses (Real Fluid)** ### **Step 1: List the Losses** - **Major (friction) losses:** For each pipe section - **Minor losses:** Entrance, contraction/expansion, exit ### **Step 2: Friction Losses Formula** \[ h_f = f \frac{L}{D} \frac{v^2}{2g} \] We'll need the **Darcy-Weisbach friction factor** \(f\). ### **Step 3: Continuity (Conservation of Mass)** \[ Q = v_1 A_1 = v_2 A_2 = v_3 A_3 \] From part (a): \(A_1 = .0452\, \text{m}^2\), \(A_2 = .00785\, \text{m}^2\), \(A_3 = .0113\, \text{m}^2\) Let \(v_3 = v\). Then, \[ v_1 = v \frac{A_3}{A_1}, \quad v_2 = v \frac{A_3}{A_2}, \quad v_3 = v \] ### **Step 4: Bernoulli's Equation (Including Losses)** \[ gH = \frac{v^2}{2} + \sum h_{loss} \] \[ gH = \frac{v^2}{2} + h_{f1} + h_{f2} + h_{f3} + h_{entrance} + h_{contraction} + h_{expansion} + h_{exit} \] #### **Minor Loss Coefficients** - Entrance (sharp): \(K_{ent} = .5\) - Sudden contraction: \(K_{contr} \approx .42 \left[1 - \left(\frac{A_2}{A_1}\right)\right]^2\) - Sudden expansion: \(K_{exp} = \left(1 - \frac{A_2}{A_1}\right)^2\) - Exit: \(K_{exit} = 1\) #### **Pipe Friction Factor** For slightly rusted steel pipe at 10°C, let's **estimate \(f = .025\)** for all sections as an average value (for turbulent flow and relative roughness). --- ### **Step 5: Write the Head Losses in Terms of \(v\)** #### **Friction Losses** \[ h_{f1} = f_1 \frac{L_1}{D_1} \frac{v_1^2}{2g} \] \[ h_{f2} = f_2 \frac{L_2}{D_2} \frac{v_2^2}{2g} \] \[ h_{f3} = f_3 \frac{L_3}{D_3} \frac{v^2}{2g} \] #### **Minor Losses** \[ h_{ent} = K_{ent} \frac{v_1^2}{2g} \] \[ h_{contr} = K_{contr} \frac{v_2^2}{2g} \] \[ h_{exp} = K_{exp} \frac{v_3^2}{2g} \] \[ h_{exit} = K_{exit} \frac{v^2}{2g} \] #### **Calculate Areas** \[ A_1 = .0452\, \text{m}^2, \; A_2 = .00785\, \text{m}^2, \; A_3 = .0113\, \text{m}^2 \] \[ \frac{A_3}{A_1} = .25, \quad \frac{A_3}{A_2} = 1.44 \] So, \[ v_1 = .25v, \quad v_2 = 1.44v, \quad v_3 = v \] #### **Plug Values into Losses** - **Friction:** - \(h_{f1} = f \frac{L_1}{D_1} \frac{(.25v)^2}{2g} = .025 \cdot \frac{3}{.24} \cdot \frac{.0625v^2}{2g} = .025 \cdot 12.5 \cdot .0625 \frac{v^2}{2g} = .01953 \frac{v^2}{2g}\) - \(h_{f2} = .025 \cdot \frac{1}{.1} \cdot (2.07v^2) \cdot \frac{1}{2g} = .025 \cdot 10 \cdot 2.07 \frac{v^2}{2g} = .5175 \frac{v^2}{2g}\) (Since \(v_2 = 1.44v \Rightarrow v_2^2 = 2.07v^2\)) - \(h_{f3} = .025 \cdot \frac{2}{.12} \cdot \frac{v^2}{2g} = .025 \cdot 16.67 \frac{v^2}{2g} = .417 \frac{v^2}{2g}\) - **Minor Losses:** - Entrance: \(h_{ent} = .5 \frac{v_1^2}{2g} = .5 \cdot .0625 \frac{v^2}{2g} = .03125 \frac{v^2}{2g}\) - Contraction: \(K_{contr} \approx .42[1 - (A_2/A_1)]^2\) - \(A_2/A_1 = .1738 \Rightarrow K_{contr} = .42(1 - .1738)^2 = .42 \cdot .682 = .286\) - \(h_{contr} = .286 \cdot (1.44^2) \frac{v^2}{2g} = .286 \cdot 2.07 \frac{v^2}{2g} = .592 \frac{v^2}{2g}\) - Expansion: \(K_{exp} = (1 - \frac{A_3}{A_2})^2 = (1 - 1.44)^2 = .1936\) - \(h_{exp} = .1936 \frac{v^2}{2g}\) - Exit: \(h_{exit} = 1 \cdot \frac{v^2}{2g}\) --- ### **Step 6: Total Head Losses** Sum all the coefficients: \[ h_{tot} = [.01953 + .5175 + .417 + .03125 + .592 + .1936 + 1] \frac{v^2}{2g} \] \[ h_{tot} = 2.7709 \frac{v^2}{2g} \] --- ### **Step 7: Full Energy Equation** \[ gH = \frac{v^2}{2} + 2.7709 \frac{v^2}{2} \] \[ gH = (1 + 2.7709)\frac{v^2}{2} \] \[ gH = 3.7709 \frac{v^2}{2} \] \[ v^2 = \frac{2gH}{3.7709} \] \[ v = \sqrt{\frac{2 \cdot 9.81 \cdot 1.5}{3.7709}} \] \[ v = \sqrt{\frac{29.43}{3.7709}} = \sqrt{7.805} = 2.793\, \text{m/s} \] \[ Q = v \cdot A_3 = 2.793 \cdot .0113 = .0316\, \text{m}^3/\text{s} \] --- ## **Final Answers** ### **a) Neglecting losses:** \[ \boxed{Q = .0613\, \text{m}^3/\text{s}} \] ### **b) Considering losses:** \[ \boxed{Q = .0316\, \text{m}^3/\text{s}} \] --- **If you need more precise values (e.g. with more accurate friction factors or more minor losses), you can recalculate, but this method is standard for undergraduate fluid mechanics.**