Let's solve the problem step by step: ### **Given Data** - **Inlet pressure**: \(P_{in} = 400 \text{ kPa (gage)}\) - **Exit pressure**: Atmospheric (let's take as reference, so \(P_{out} = 0\)) - **Neglect minor losses** - **Pipe Data** (all lengths \(L\) in m, diameters \(D\) in mm): - \(L_0 = 400, D_0 = 75\) - \(L_1 = 300, D_1 = 50\) - \(L_2 = 150, D_2 = 50\) - \(L_3 = 150, D_3 = 35\) - \(L_4 = 100, D_4 = 75\) - **Find:** \(Q_0, Q_1, Q_2, Q_3, Q_4\) ### **Step 1: Continuity (Mass Conservation)** At the junctions, flow rates must balance: - At inlet: \(Q_0 = Q_1 + Q_2 + Q_3 + Q_4\) - At the T-junction: - First branch: \(Q_1\) (right) - Second branch splits: \(Q_2\) (right), \(Q_3\) (down), \(Q_4\) (left) ### **Step 2: Head Losses (Darcy-Weisbach Equation)** For each pipe: \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] Where \(h_f\) = head loss, \(f\) = friction factor, \(L\) = length, \(D\) = diameter, \(V\) = velocity, \(g\) = 9.81 m/s². For water, assume **density \(\rho = 1000 \, \text{kg/m}^3\)** and **kinematic viscosity \(\nu \approx 1 \times 10^{-6} \, \text{m}^2/\text{s}\)**. ### **Step 3: Flow Relationships** Let’s assign: - \(Q_0\) entering node, splits into \(Q_1, Q_2, Q_3, Q_4\). - All end at atmospheric pressure. #### **Write energy equations from inlet to each outlet:** \[ \frac{P_{in}}{\gamma} = h_{f,0} + h_{f,1} \] (for branch 1, and similarly for other branches.) #### **Convert all diameters to meters:** - \(D_0 = 0.075\,\text{m}\) - \(D_1 = 0.050\,\text{m}\) - \(D_2 = 0.050\,\text{m}\) - \(D_3 = 0.035\,\text{m}\) - \(D_4 = 0.075\,\text{m}\) #### **Flow Areas:** \[ A = \frac{\pi D^2}{4} \] So, velocity in a pipe: \(V = Q/A\). #### **Let’s Write the Equations:** **Let’s define node 0 as inlet, and the branches as:** - \(Q_0\) into system - \(Q_1\) out right - \(Q_2\) out right from branch - \(Q_3\) out down - \(Q_4\) out left #### **Energy Equation (from inlet to each outlet):** \[ \frac{P_{in}}{\gamma} = \sum h_f \quad \text{(along each path)} \] Where \(\gamma = \rho g\): \[ \frac{P_{in}}{\gamma} = \frac{400,000}{1000 \times 9.81} = 40.77\,\text{m} \] --- ### **Write losses along each path:** #### **Path 1 (to \(Q_1\)):** \[ h_{f,0} + h_{f,1} = f_0 \frac{L_0}{D_0} \frac{V_0^2}{2g} + f_1 \frac{L_1}{D_1} \frac{V_1^2}{2g} \] #### **Path 2 (to \(Q_2\)):** \[ h_{f,0} + h_{f,4} + h_{f,2} = f_0 \frac{L_0}{D_0} \frac{V_0^2}{2g} + f_4 \frac{L_4}{D_4} \frac{V_4^2}{2g} + f_2 \frac{L_2}{D_2} \frac{V_2^2}{2g} \] #### **Path 3 (to \(Q_3\)):** \[ h_{f,0} + h_{f,4} + h_{f,3} = f_0 \frac{L_0}{D_0} \frac{V_0^2}{2g} + f_4 \frac{L_4}{D_4} \frac{V_4^2}{2g} + f_3 \frac{L_3}{D_3} \frac{V_3^2}{2g} \] #### **Path 4 (to \(Q_4\)):** \[ h_{f,0} + h_{f,4} = f_0 \frac{L_0}{D_0} \frac{V_0^2}{2g} + f_4 \frac{L_4}{D_4} \frac{V_4^2}{2g} \] --- ### **Step 4: Assume All Pipes Have Same Roughness** For galvanized iron, assume \(f \approx 0.02\) (for turbulent flow, check with Reynolds number later). --- ### **Step 5: Express All Variables in Terms of \(Q\)** \[ V = \frac{Q}{A} \] --- ### **Step 6: Set Up System of Equations** Let’s define: - \(Q_1\) through \(L_1, D_1\) - \(Q_2\) through \(L_2, D_2\) - \(Q_3\) through \(L_3, D_3\) - \(Q_4\) through \(L_4, D_4\) - \(Q_0\) through \(L_0, D_0\) **At the junction:** \[ Q_0 = Q_1 + Q_2 + Q_3 + Q_4 \] --- ### **Write All Equations (with f = 0.02):** Let’s denote \(A_i\) for each diameter: - \(A_0 = \pi/4 \times (0.075)^2 = 0.00442\,\text{m}^2\) - \(A_1 = \pi/4 \times (0.05)^2 = 0.00196\,\text{m}^2\) - \(A_2 = A_1\) - \(A_3 = \pi/4 \times (0.035)^2 = 0.000962\,\text{m}^2\) - \(A_4 = A_0\) Velocities: - \(V_i = Q_i / A_i\) #### **Four Energy Equations:** (1) **To \(Q_1\):** \[ 40.77 = 0.02 \frac{400}{0.075} \frac{(Q_0/A_0)^2}{2 \times 9.81} + 0.02 \frac{300}{0.05} \frac{(Q_1/A_1)^2}{2 \times 9.81} \] (2) **To \(Q_2\):** \[ 40.77 = 0.02 \frac{400}{0.075} \frac{(Q_0/A_0)^2}{2 \times 9.81} + 0.02 \frac{100}{0.075} \frac{(Q_4/A_4)^2}{2 \times 9.81} + 0.02 \frac{150}{0.05} \frac{(Q_2/A_2)^2}{2 \times 9.81} \] (3) **To \(Q_3\):** \[ 40.77 = 0.02 \frac{400}{0.075} \frac{(Q_0/A_0)^2}{2 \times 9.81} + 0.02 \frac{100}{0.075} \frac{(Q_4/A_4)^2}{2 \times 9.81} + 0.02 \frac{150}{0.035} \frac{(Q_3/A_3)^2}{2 \times 9.81} \] (4) **To \(Q_4\):** \[ 40.77 = 0.02 \frac{400}{0.075} \frac{(Q_0/A_0)^2}{2 \times 9.81} + 0.02 \frac{100}{0.075} \frac{(Q_4/A_4)^2}{2 \times 9.81} \] --- ### **Step 7: Use Continuity at the Junction** \[ Q_0 = Q_1 + Q_2 + Q_3 + Q_4 \] --- ### **Step 8: Solve System Numerically** Due to the nonlinearity (squared terms), this system is best solved numerically (e.g., using MATLAB, Python, or trial and error). But let's create a simplified approach: #### **Guess values for \(Q_4\), then solve for others:** From equation (4) (to \(Q_4\)): \[ 40.77 = 0.02 \frac{400}{0.075} \frac{(Q_0/A_0)^2}{2 \times 9.81} + 0.02 \frac{100}{0.075} \frac{(Q_4/A_4)^2}{2 \times 9.81} \] But \(Q_0 = Q_1 + Q_2 + Q_3 + Q_4\) --- #### **You can use Excel or Python to solve these, but let’s make a rough estimation:** Suppose each branch takes approximately the same share (since all outlets at atmospheric pressure): Try \(Q_1 \approx Q_2 \approx Q_3 \approx Q_4\). Let \(Q_1 = Q_2 = Q_3 = Q_4 = x\), so \(Q_0 = 4x\). Substitute into equation for \(Q_4\): \[ 40.77 = 0.02 \frac{400}{0.075} \frac{(4x/0.00442)^2}{2 \times 9.81} + 0.02 \frac{100}{0.075} \frac{(x/0.00442)^2}{2 \times 9.81} \] Calculate: - \(\frac{400}{0.075} = 5333\) - \(\frac{100}{0.075} = 1333\) - \(A_0 = 0.00442\) - \((4x/0.00442)^2 = (4x)^2/(0.00442)^2 = 16x^2/1.953 \times 10^{-5}\) - \((x/0.00442)^2 = x^2/1.953 \times 10^{-5}\) Plugging in values and simplifying, \[ 40.77 = 0.02 \cdot 5333 \cdot \frac{16x^2}{2 \cdot 9.81 \cdot 1.953 \times 10^{-5}} + 0.02 \cdot 1333 \cdot \frac{x^2}{2 \cdot 9.81 \cdot 1.953 \times 10^{-5}} \] Let's compute the constants: - \(2 \cdot 9.81 \cdot 1.953 \times 10^{-5} \approx 0.0003837\) So, \[ 0.02 \cdot 5333 \cdot \frac{16x^2}{0.0003837} = 0.02 \cdot 5333 \cdot 41708x^2 \] \(= 0.02 \cdot 222,708,000 x^2 \approx 4,454,160 x^2\) Similarly, \(0.02 \cdot 1333 \cdot \frac{x^2}{0.0003837} = 0.02 \cdot 1333 \cdot 2606 x^2 \approx 69,420 x^2\) So, \[ 40.77 = 4,454,160 x^2 + 69,420 x^2 = 4,523,580 x^2 \] \[ x^2 = \frac{40.77}{4,523,580} \] \[ x^2 = 9.02 \times 10^{-6} \] \[ x = 0.003 \text{ m}^3/\text