Let's solve the problem step by step. **Given:** - Distance from tower to subdivision, \(L = 2\) miles \(= 10,560\) ft - Elevation difference, \(\Delta z = 150\) ft (tower above ground) - Flow rate, \(Q = 10,000\) gpm - Pressure at subdivision, \(P_2 = 30\) psig - Pressure at top of water in tank, \(P_1 = 1\) atm (assume vented to atmosphere) - Schedule 40 commercial steel pipe (we'll use roughness \(\epsilon = 0.00015\) ft) - Ignore kinetic energy change (i.e., \(\Delta V^2/2g \approx 0\)) - Water temperature ~room temperature, so \(\rho = 62.4\) lb/ft³, \(\mu = 1.1 \times 10^{-5}\) lb·s/ft² --- ## **Step 1: Write the Energy Equation (Bernoulli with head loss)** Between the water surface in the tank (point 1) and the outlet in the subdivision (point 2): \[ \frac{P_1}{\gamma} + z_1 = \frac{P_2}{\gamma} + z_2 + h_f \] Where: - \(P_1 =\) atmospheric pressure (gauge = 0 psig) - \(z_1 = 150\) ft - \(P_2 = 30\) psig \(= 30 + 14.7 = 44.7\) psia (but we will use gauge, so just 30 psig) - \(z_2 = 0\) ft (street level) - \(h_f =\) head loss in pipe \(\gamma = 62.4\) lb/ft³ Rearrange: \[ h_f = \frac{P_1}{\gamma} + z_1 - \left(\frac{P_2}{\gamma} + z_2\right) \] Because \(P_1\) is atmospheric, \(P_1 = 0\) psig, so \(\frac{P_1}{\gamma} = 0\): \[ h_f = 150 - \frac{30 \times 144}{62.4} \] *Note: 1 psi = 144 psf; convert 30 psi to lb/ft²: 30 × 144 = 4320 lb/ft².* \[ \frac{4320}{62.4} = 69.23\ \text{ft} \] So, \[ h_f = 150 - 69.23 = 80.77\ \text{ft} \] --- ## **Step 2: Head Loss in Pipe (Darcy-Weisbach Equation)** \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] Where: - \(f =\) friction factor - \(L = 10,560\) ft - \(D =\) pipe diameter (ft) - \(V =\) average velocity (ft/s) - \(g = 32.2\) ft/s² --- ## **Step 3: Flow Rate and Velocity** \[ Q = AV \implies V = \frac{Q}{A} = \frac{Q}{\frac{\pi}{4} D^2} \] Convert Q to ft³/s: \[ Q = 10,000\ \text{gpm} \times \frac{1\ \text{ft}^3}{7.48\ \text{gal}} \times \frac{1}{60\ \text{s/min}} = \frac{10,000}{7.48 \times 60} = 22.29\ \text{ft}^3/\text{s} \] --- ## **Step 4: Substitute and Rearrange** Plug in \(V = \frac{4Q}{\pi D^2}\): \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] \[ 80.77 = f \cdot \frac{10,560}{D} \cdot \frac{1}{2 \times 32.2} \left( \frac{4Q}{\pi D^2} \right)^2 \] \[ = f \cdot \frac{10,560}{D} \cdot \frac{1}{64.4} \cdot \frac{16Q^2}{\pi^2 D^4} \] \[ = f \cdot \frac{10,560}{D} \cdot \frac{16Q^2}{64.4 \pi^2 D^4} \] \[ = f \cdot \frac{10,560 \times 16 Q^2}{64.4 \pi^2 D^5} \] \[ 80.77 = f \cdot \frac{10,560 \times 16 \times (22.29)^2}{64.4 \pi^2 D^5} \] Calculate \(16 \times (22.29)^2\): \[ (22.29)^2 = 497 \] \[ 16 \times 497 = 7,952 \] So, \[ 80.77 = f \cdot \frac{10,560 \times 7,952}{64.4 \pi^2 D^5} \] Calculate numerator: \(10,560 \times 7,952 = 83,990,720\) Calculate denominator: \(64.4 \times \pi^2 = 64.4 \times 9.87 = 635.7\) \[ 80.77 = f \cdot \frac{83,990,720}{635.7 \cdot D^5} \] \[ 80.77 = f \cdot \frac{132,205}{D^5} \] \[ 80.77 D^5 = f \cdot 132,205 \] \[ D^5 = \frac{f \cdot 132,205}{80.77} \] \[ D^5 = 1,636.36 f \] --- ## **Step 5: Estimate f and Solve for D** For commercial steel pipe, \(\epsilon = 0.00015\) ft. The flow is turbulent, but initial guess: for large pipes, \(f \approx 0.018 - 0.022\). Let’s iterate: **First, assume \(f = 0.02\)**: \[ D^5 = 1,636.36 \times 0.02 = 32.73 \] \[ D = (32.73)^{1/5} = \exp\left(\frac{1}{5} \ln 32.73\right) \] \[ \ln 32.73 = 3.49 \] \[ \frac{3.49}{5} = 0.698 \] \[ e^{0.698} = 2.01 \text{ ft} \] **So, D ≈ 2.01 ft = 24.1 inches** --- ## **Step 6: Calculate Velocity and Check Reynolds Number** \[ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (2.01)^2 = \frac{\pi}{4} \times 4.04 = 3.18\ \text{ft}^2 \] \[ V = \frac{Q}{A} = \frac{22.29}{3.18} = 7.01\ \text{ft/s} \] Reynolds number: \[ Re = \frac{VD}{\nu} \] Kinematic viscosity of water: \(\nu = 1.1 \times 10^{-5} \frac{\text{ft}^2}{\text{s}}\) \[ Re = \frac{7.01 \times 2.01}{1.1 \times 10^{-5}} = \frac{14.09}{1.1 \times 10^{-5}} \] \[ = 1.28 \times 10^6 \] --- ## **Step 7: Calculate f Using Colebrook-White (or Swamee-Jain)** Swamee-Jain: \[ f = 0.25 \left[ \log_{10} \left( \frac{\epsilon}{3.7D} + \frac{5.74}{Re^{0.9}} \right) \right]^{-2} \] \[ \frac{\epsilon}{3.7D} = \frac{0.00015}{3.7 \times 2.01} = \frac{0.00015}{7.437} = 2.02 \times 10^{-5} \] \[ \frac{5.74}{Re^{0.9}} = \frac{5.74}{(1.28 \times 10^6)^{0.9}} \] Take log: \[ \log_{10} (1.28 \times 10^6) = 6.107 \] \[ (1.28 \times 10^6)^{0.9} = 10^{6.107 \times 0.9} = 10^{5.50} = 3.16 \times 10^5 \] \[ \frac{5.74}{3.16 \times 10^5} = 1.82 \times 10^{-5} \] Sum: \[ 2.02 \times 10^{-5} + 1.82 \times 10^{-5} = 3.84 \times 10^{-5} \] \[ \log_{10} (3.84 \times 10^{-5}) = \log_{10}(3.84) - 5 = 0.584 - 5 = -4.416 \] \[ f = 0.25 \left[ -4.416 \right]^{-2} = 0.25 / (19.51) = 0.0128 \] --- ## **Step 8: Recalculate D Using New f** \[ D^5 = 1,636.36 \times 0.0128 = 20.95 \] \[ D = (20.95)^{1/5} \] \[ \ln 20.95 = 3.044; \ 3.044/5 = 0.609; \ e^{0.609} = 1.84\ \text{ft} = 22.1\ \text{in} \] --- ## **Step 9: Iterate Again** Update the velocity: \[ A = \frac{\pi}{4} (1.84)^2 = \frac{\pi}{4} \times 3.39 = 2.66\ \text{ft}^2 \] \[ V = \frac{22.29}{2.66} = 8.38\ \text{ft/s} \] \[ Re = \frac{8.38 \times 1.84}{1.1 \times 10^{-5}} = \frac{15.42}{1.1 \times 10^{-5}} = 1.40 \times 10^6 \] Swamee-Jain: \[ \frac{\epsilon}{3.7D} = \frac{0.00015}{3.7 \times 1.84} = 2.19 \times 10^{-5} \] \[ \frac{5.74}{(1.4 \times 10^6)^{0.9}} = \frac{5.74}{(10^{6.146 \times 0.9})} = 10^{5.53} = 3.39 \times 10^5 \] \[ \frac{5.74}{3.39 \times 10^5} = 1.69 \times 10^{-5} \] Sum: \(2.19 + 1.69 = 3.88 \times 10^{-5}\) \[ \log_{10} (3.88 \times 10^{-5}) = -4.411 \] \[ f = 0.25 / (19.45) = 0.01285 \] So f is essentially the same. Thus, **converged**. --- ## **Step 10: Standard Pipe Size** Actual inside diameters for Schedule 40 pipe: - 20 in: 19.00 in = 1.583 ft - 22 in: 21.00 in = 1.75 ft - 24 in: 22.50 in = 1.875 ft Our calculated D is 1.84 ft (22.1 in), so **use 24 inch Schedule 40** pipe (ID = 22.5 in = 1.875 ft). --- ## **Step 11: Check Assumption (Neglecting Kinetic Energy Term)** Kinetic energy term: \[ \Delta \left(\frac{V^2}{2g}\right) = \frac{V_2^2 - V_1^2}{2g} \] Assume V1 (water surface in tank) ≈ 0, so \[ \frac{V_2^2}{2g} = \frac{(8.38)^2}{2 \times 32.2} = \frac{70.2}{64.4} = 1.09\ \text{ft} \]