1. **Problem statement:** Calculate the power required to pump aqueous nitric acid through a steel pipe, the gauge pressure at the pump outlet, and the effects of increased pipe roughness on power and pressure. 2. **Given data:** - Pipe length, $L = 278$ m - Pipe diameter, $D = 3.20$ cm = 0.032 m - Initial absolute roughness, $\epsilon_1 = 0.0035$ mm = 3.5 \times 10^{-6}$ m - Corroded roughness, $\epsilon_2 = 0.050$ mm = 5.0 \times 10^{-5}$ m - Vertical discharge height, $z = 57.4$ m - Suction depth below liquid surface, $h_s = 5.60$ m - Density, $\rho = 1068$ kg/m$^3$ - Viscosity, $\mu = 1.06 \times 10^{-3}$ Pa\cdots - Mass flow rate, $\dot{m} = 2.35$ kg/s 3. **Step 1: Calculate volumetric flow rate $Q$ and velocity $V$** $$Q = \frac{\dot{m}}{\rho} = \frac{2.35}{1068} = 0.0022\, \text{m}^3/\text{s}$$ Cross-sectional area: $$A = \frac{\pi D^2}{4} = \frac{\pi (0.032)^2}{4} = 8.042 \times 10^{-4} \text{ m}^2$$ Velocity: $$V = \frac{Q}{A} = \frac{0.0022}{8.042 \times 10^{-4}} = 2.74 \text{ m/s}$$ 4. **Step 2: Calculate Reynolds number $Re$** $$Re = \frac{\rho V D}{\mu} = \frac{1068 \times 2.74 \times 0.032}{1.06 \times 10^{-3}} = 8.83 \times 10^{4}$$ 5. **Step 3: Calculate friction factor $f$ using Colebrook-White equation approximation** Relative roughness: $$\frac{\epsilon_1}{D} = \frac{3.5 \times 10^{-6}}{0.032} = 1.09 \times 10^{-4}$$ Using the Swamee-Jain equation for turbulent flow: $$f = 0.25 \left[ \log_{10} \left( \frac{\epsilon_1}{3.7D} + \frac{5.74}{Re^{0.9}} \right) \right]^{-2}$$ Calculate inside log: $$\frac{\epsilon_1}{3.7D} = \frac{3.5 \times 10^{-6}}{3.7 \times 0.032} = 2.95 \times 10^{-5}$$ $$\frac{5.74}{Re^{0.9}} = \frac{5.74}{(8.83 \times 10^{4})^{0.9}} = 1.22 \times 10^{-4}$$ Sum: $$1.51 \times 10^{-4}$$ Log base 10: $$\log_{10}(1.51 \times 10^{-4}) = -3.82$$ Friction factor: $$f = 0.25 \times (-3.82)^{-2} = 0.0171$$ 6. **Step 4: Calculate head loss due to friction $h_f$** $$h_f = f \frac{L}{D} \frac{V^2}{2g}$$ Where $g = 9.81$ m/s$^2$ $$h_f = 0.0171 \times \frac{278}{0.032} \times \frac{(2.74)^2}{2 \times 9.81} = 20.9 \text{ m}$$ 7. **Step 5: Calculate total head $H$ the pump must provide** Suction lift is 5.60 m below surface, so total vertical lift: $$H = h_f + z + h_s = 20.9 + 57.4 + 5.6 = 83.9 \text{ m}$$ 8. **Step 6: Calculate power required $P$** $$P = \rho g Q H = 1068 \times 9.81 \times 0.0022 \times 83.9 = 1920 \text{ W} = 1.92 \text{ kW}$$ 9. **Step 7: Calculate gauge pressure at pump outlet $P_g$** Using Bernoulli and head loss: $$P_g = \rho g (h_f + h_s) = 1068 \times 9.81 \times (20.9 + 5.6) = 273,000 \text{ Pa} = 273 \text{ kPa}$$ 10. **Step 8: Repeat friction factor calculation for corroded roughness $\epsilon_2$** $$\frac{\epsilon_2}{D} = \frac{5.0 \times 10^{-5}}{0.032} = 0.00156$$ Inside log: $$\frac{\epsilon_2}{3.7D} = \frac{5.0 \times 10^{-5}}{3.7 \times 0.032} = 4.22 \times 10^{-4}$$ Sum inside log: $$4.22 \times 10^{-4} + 1.22 \times 10^{-4} = 5.44 \times 10^{-4}$$ Log base 10: $$\log_{10}(5.44 \times 10^{-4}) = -3.26$$ Friction factor: $$f = 0.25 \times (-3.26)^{-2} = 0.0235$$ 11. **Step 9: Calculate new head loss $h_f'$** $$h_f' = 0.0235 \times \frac{278}{0.032} \times \frac{(2.74)^2}{2 \times 9.81} = 28.7 \text{ m}$$ 12. **Step 10: Calculate new total head $H'$ and power $P'$** $$H' = h_f' + z + h_s = 28.7 + 57.4 + 5.6 = 91.7 \text{ m}$$ $$P' = 1068 \times 9.81 \times 0.0022 \times 91.7 = 2097 \text{ W} = 2.10 \text{ kW}$$ 13. **Step 11: Calculate percentage increase in power** $$\% \text{ increase} = \frac{P' - P}{P} \times 100 = \frac{2097 - 1920}{1920} \times 100 = 9.3\%$$ 14. **Step 12: Calculate new gauge pressure $P_g'$** $$P_g' = \rho g (h_f' + h_s) = 1068 \times 9.81 \times (28.7 + 5.6) = 333,000 \text{ Pa} = 333 \text{ kPa}$$ **Final answers:** - (a) Power required = 1.92 kW - (b) Gauge pressure at pump outlet = 273 kPa - (c) Percentage increase in power due to corrosion = 9.3% - (d) Gauge pressure with corrosion = 333 kPa