1. **Problem Statement:** Calculate the power required to pump aqueous nitric acid through a steel pipe with given dimensions and conditions. 2. **Given Data:** - Length of pipe, $L = 278$ m - Diameter of pipe, $d = 3.20$ cm = 0.032 m - Absolute roughness, $\epsilon = 0.0035$ mm = 0.0000035 m - Elevation difference, $\Delta z = 57.4$ m (discharge point above pump outlet) - Suction lift, $h_s = 5.60$ m (pump draws acid from 5.60 m below tank surface) - Density, $\rho = 1068$ kg/m$^3$ - Viscosity, $\mu = 1.06 \times 10^{-3}$ Pa\cdot s - Mass flow rate, $\dot{m} = 2.35$ kg/s 3. **Step 1: Calculate volumetric flow rate $Q$** $$Q = \frac{\dot{m}}{\rho} = \frac{2.35}{1068} = 0.0022\ \text{m}^3/\text{s}$$ 4. **Step 2: Calculate flow velocity $V$** Cross-sectional area of pipe: $$A = \frac{\pi d^2}{4} = \frac{\pi (0.032)^2}{4} = 8.042 \times 10^{-4}\ \text{m}^2$$ Velocity: $$V = \frac{Q}{A} = \frac{0.0022}{8.042 \times 10^{-4}} = 2.73\ \text{m/s}$$ 5. **Step 3: Calculate Reynolds number $Re$** $$Re = \frac{\rho V d}{\mu} = \frac{1068 \times 2.73 \times 0.032}{1.06 \times 10^{-3}} = 8.79 \times 10^{4}$$ 6. **Step 4: Calculate relative roughness $\frac{\epsilon}{d}$** $$\frac{\epsilon}{d} = \frac{0.0000035}{0.032} = 1.09 \times 10^{-4}$$ 7. **Step 5: Calculate Darcy friction factor $f$ using Colebrook-White equation approximation** For turbulent flow and given roughness: $$\frac{1}{\sqrt{f}} = -2 \log_{10} \left( \frac{\epsilon}{3.7 d} + \frac{2.51}{Re \sqrt{f}} \right)$$ Using iterative or approximate formula (Swamee-Jain): $$f = 0.25 \left[ \log_{10} \left( \frac{\epsilon}{3.7 d} + \frac{5.74}{Re^{0.9}} \right) \right]^{-2}$$ Calculate inside log: $$\frac{\epsilon}{3.7 d} = \frac{0.0000035}{3.7 \times 0.032} = 2.95 \times 10^{-5}$$ $$\frac{5.74}{Re^{0.9}} = \frac{5.74}{(8.79 \times 10^{4})^{0.9}} = 1.03 \times 10^{-4}$$ Sum: $$2.95 \times 10^{-5} + 1.03 \times 10^{-4} = 1.32 \times 10^{-4}$$ Log term: $$\log_{10}(1.32 \times 10^{-4}) = -3.88$$ Friction factor: $$f = 0.25 \times (-3.88)^{-2} = 0.25 \times 0.0664 = 0.0166$$ 8. **Step 6: Calculate head loss due to friction $h_f$** $$h_f = f \frac{L}{d} \frac{V^2}{2g}$$ Where $g = 9.81$ m/s$^2$ Calculate: $$\frac{L}{d} = \frac{278}{0.032} = 8687.5$$ $$\frac{V^2}{2g} = \frac{(2.73)^2}{2 \times 9.81} = 0.38\ \text{m}$$ So, $$h_f = 0.0166 \times 8687.5 \times 0.38 = 54.8\ \text{m}$$ 9. **Step 7: Calculate total head $H$ the pump must overcome** Total head includes elevation difference, suction lift, and friction loss: $$H = \Delta z + h_s + h_f = 57.4 + 5.6 + 54.8 = 117.8\ \text{m}$$ 10. **Step 8: Calculate power required $P$** Power is given by: $$P = \rho g Q H$$ Calculate: $$P = 1068 \times 9.81 \times 0.0022 \times 117.8 = 2707\ \text{W} = 2.71\ \text{kW}$$ **Final answer:** The power required to deliver the acid at the given rate is approximately **2.71 kW**.