Let's break down the solution step by step. --- ## **Given Data** - **Pressure difference (ΔP):** 25 psi - **Pipes are galvanized iron:** \( k_s = .0005 \) ft - **Kinematic viscosity (\( \nu \)):** \( 1.06 \times 10^{-5} \) ft²/s - **Neglect minor losses** - **Pipe layout:** - 10" dia (inlet, 1 to A) - 200 ft, 8" dia (A) - 160 ft, 6" dia (B) - 800 ft, 10" dia (C to 2) --- ## **Step 1: Convert Pressure Difference to Feet of Water** \[ \Delta P = 25 \text{ psi} \times \frac{144 \text{ in}^2}{1 \text{ ft}^2} = 360 \text{ psf} \] \[ \text{Head difference:} \] \[ \Delta h = \frac{\Delta P}{\gamma} \] where \( \gamma = 62.4 \) lb/ft³ (unit weight of water) \[ \Delta h = \frac{25 \times 144}{62.4} = \frac{360}{62.4} = 57.69 \text{ ft} \] --- ## **Step 2: Identify Flow Path and Equivalent Pipe** This is a parallel pipe system (A and B are in parallel). - Path A: 200 ft, 8" dia - Path B: 160 ft, 6" dia The total head loss from 1 to 2 is the sum of head losses in the 10" sections and the parallel section (either A or B). Let \( Q \) = total flowrate, \( Q_A \) = flow in A, \( Q_B \) = flow in B \[ Q = Q_A + Q_B \] --- ## **Step 3: Head Loss Equations (Darcy-Weisbach)** \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] Let’s denote: - Section 1 (10" inlet): \( L_1 = 10 \) ft, \( D_1 = 10 \) in = .833 ft - Section A: \( L_A = 200 \) ft, \( D_A = 8 \) in = .667 ft - Section B: \( L_B = 160 \) ft, \( D_B = 6 \) in = .5 ft - Section C (10" outlet): \( L_C = 800 \) ft, \( D_C = 10 \) in = .833 ft --- ## **Step 4: Assume Velocity and Flow Variables** Let \( Q_A \) and \( Q_B \) be the flow rates in pipes A and B. The velocity in each pipe is: \[ V = \frac{Q}{A} \] Where \( A = \frac{\pi}{4} D^2 \) --- ## **Step 5: Head Loss in Each Path** ### **For the 10" pipes (inlet + outlet):** Total length: \( L_{10} = 10 + 800 = 810 \) ft, \( D_{10} = .833 \) ft \[ h_{10} = f_{10} \frac{810}{.833} \frac{V_{10}^2}{2g} \] ### **For the parallel section:** The head loss through A and B must be equal: \[ h_A = h_B \] \[ h_A = f_A \frac{200}{.667} \frac{V_A^2}{2g} \] \[ h_B = f_B \frac{160}{.5} \frac{V_B^2}{2g} \] --- ## **Step 6: Express Everything in Terms of Q** Let: - \( Q = Q_A + Q_B \) - \( Q_A = V_A \cdot A_A \), \( Q_B = V_B \cdot A_B \) - \( Q = V_{10} \cdot A_{10} \) Areas: - \( A_{10} = \frac{\pi}{4}(.833^2) = .545 \) ft² - \( A_A = \frac{\pi}{4}(.667^2) = .349 \) ft² - \( A_B = \frac{\pi}{4}(.5^2) = .196 \) ft² --- ## **Step 7: Find f (Darcy friction factor) for Each Pipe** First, estimate \( f \) using the Colebrook-White equation (or use Swamee-Jain for practical calculations): \[ f = .25 \left[ \log_{10}\left( \frac{k_s}{3.7D} + \frac{5.74}{Re^{.9}} \right) \right]^{-2} \] where \( Re = \frac{VD}{\nu} \) This will require iteration because \( f \) depends on velocity, which depends on \( Q \). --- ## **Step 8: Set Up the Equations** ### **Total Head Loss from 1 to 2:** \[ \Delta h = h_{10} + h_{parallel} \] Since the head loss through the parallel pipe is the same for both A and B, and the total flow is distributed: \[ h_{parallel} = h_A = h_B \] \[ \Delta h = f_{10} \frac{810}{.833} \frac{V_{10}^2}{2g} + h_{parallel} \] --- ## **Step 9: Iterative Solution** Let's **assume \( f \approx .02 \)** for initial calculations (typical for water in galvanized pipes). ### **Let’s Guess a Total Flow \( Q \) and Iterate** Let’s denote: \[ Q = Q_A + Q_B \] Let’s try \( Q = 5 \) cfs as a guess. #### **Find Velocities:** \[ V_{10} = \frac{Q}{A_{10}} = \frac{5}{.545} = 9.17 \text{ ft/s} \] \[ V_A = \frac{Q_A}{A_A}, \quad V_B = \frac{Q_B}{A_B} \] #### **Head Loss in Parallel Pipes:** \[ h_A = f_A \frac{200}{.667} \frac{V_A^2}{2g} \] \[ h_B = f_B \frac{160}{.5} \frac{V_B^2}{2g} \] Set \( h_A = h_B \): \[ f_A \frac{200}{.667} \frac{V_A^2}{2g} = f_B \frac{160}{.5} \frac{V_B^2}{2g} \] Assuming \( f_A \approx f_B \): \[ \frac{200}{.667} V_A^2 = \frac{160}{.5} V_B^2 \] \[ 2999 V_A^2 = 320 V_B^2 \] \[ V_A^2 / V_B^2 = 320 / 2999 \implies V_A / V_B = \sqrt{1.067} = 1.033 \] But also, \( Q_A + Q_B = Q \): \[ Q_A = V_A \cdot A_A, \quad Q_B = V_B \cdot A_B \] \[ Q = V_A \cdot .349 + V_B \cdot .196 \] \[ V_A = 1.033 V_B \] \[ Q = .349 \times 1.033 V_B + .196 V_B = (.360 + .196)V_B = .556 V_B \] \[ V_B = Q / .556 \] \[ V_A = 1.033 V_B = 1.033 \times Q / .556 \] For \( Q = 5 \): \[ V_B = 5 / .556 = 8.99 \text{ ft/s} \] \[ V_A = 1.033 \times 8.99 = 9.28 \text{ ft/s} \] --- #### **Calculate Head Losses** ##### **In 10" Pipe (inlet + outlet):** \[ h_{10} = .02 \times \frac{810}{.833} \times \frac{(9.17)^2}{2 \times 32.2} \] \[ \frac{810}{.833} = 972.9 \] \[ (9.17)^2 = 84.08 \] \[ h_{10} = .02 \times 972.9 \times \frac{84.08}{64.4} \] \[ = .02 \times 972.9 \times 1.306 \] \[ = .02 \times 1271.6 = 25.43 \text{ ft} \] ##### **In Parallel Path (A or B):** Let’s use A: \[ h_A = .02 \times \frac{200}{.667} \times \frac{(9.28)^2}{2 \times 32.2} \] \[ \frac{200}{.667} = 2999 \] \[ (9.28)^2 = 86.14 \] \[ h_A = .02 \times 2999 \times \frac{86.14}{64.4} \] \[ = .02 \times 2999 \times 1.338 \] \[ = .02 \times 4013.7 = 80.27 \text{ ft} \] ##### **Total Head Loss:** \[ h_{total} = h_{10} + h_A = 25.43 + 80.27 = 105.7 \text{ ft} \] But the available head is only 57.69 ft. --- ## **Step 10: Adjust Flow Rate and Iterate** Since calculated \( h_{total} > 57.69 \), our guess for \( Q \) is too high. Try \( Q = 3 \) cfs: \[ V_{10} = 3 / .545 = 5.5 \text{ ft/s} \] \[ V_B = 3 / .556 = 5.4 \text{ ft/s} \] \[ V_A = 1.033 \times 5.4 = 5.58 \text{ ft/s} \] #### **Head Losses:** - \( h_{10} = .02 \times 972.9 \times \frac{(5.5)^2}{64.4} = .02 \times 972.9 \times .470 = .02 \times 457.3 = 9.15 \text{ ft} \) - \( h_A = .02 \times 2999 \times \frac{(5.58)^2}{64.4} = .02 \times 2999 \times .484 = .02 \times 145.7 = 29.01 \text{ ft} \) - \( h_{total} = 9.15 + 29.01 = 38.16 \text{ ft} \) Now \( h_{total} < 57.69 \), so \( Q \) can be higher. Try \( Q = 4 \) cfs: - \( V_{10} = 4/.545 = 7.34 \text{ ft/s} \) - \( V_B = 4/.556 = 7.19 \) - \( V_A = 1.033 \times 7.19 = 7.43 \) - \( h_{10} = .02 \times 972.9 \times (7.34^2 / 64.4) = .02 \times 972.9 \times .837 = .02 \times 814.4 = 16.29 \) - \( h_A = .02 \times 2999 \times (7.43^2 / 64.4) = .02 \times 2999 \times .857 = .02 \times 257.7 = 51.41 \) - \( h_{total} = 16.29 + 51.41 = 67.70 \text{ ft} \) Now \( h_{total} > 57.69 \), so \( Q \) is between 3 and 4 cfs. --- ## **Step 11: Linear Interpolation** Let’s interpolate between \( Q = 3 \) (38.16 ft) and \( Q = 4 \) (67.70 ft): \[ Q = 3 + \frac{57.69 - 38.16}{67.70 - 38.16} \times (4-3) \] \[ Q = 3 + \frac{19.53}{