Let's break down the problem step by step: ### **Given Data** - **Flow rate (\(\dot{m}\))**: \(60\) ton/hr = \(60,000\) kg/hr = \(16.67\) kg/s - **Pipe Nominal Size**: 80 mm (Assume actual internal diameter for Sch 40 is about **ID = 77.9 mm = 0.0779 m**) - **Pipe length (L)**: 250 m - **Miscellaneous losses**: Equivalent to 500 pipe diameters - **Density (\(\rho\))**: 950 kg/m³ - **Viscosity (\(\mu\))**: 1.0 mN·m⁻²·s = 1.0 cP = 0.001 N·s/m² - **Tank pressure**: Atmospheric = 1.013 bar = 101,300 Pa - **Column pressure**: 2 bara = 200,000 Pa - **Tank level above pump**: 2 m - **Column feed point above pump**: 6 m - **Pump efficiency (\(\eta\))**: 70% = 0.7 --- ## **Step 1: Calculate Volumetric Flow Rate (\(Q\))** \[ Q = \frac{\dot{m}}{\rho} = \frac{16.67\ \text{kg/s}}{950\ \text{kg/m}^3} = 0.01755\ \text{m}^3/\text{s} \] --- ## **Step 2: Calculate Pipe Velocity (\(v\))** \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.0779)^2 = 0.00476\ \text{m}^2 \] \[ v = \frac{Q}{A} = \frac{0.01755}{0.00476} = 3.69\ \text{m/s} \] --- ## **Step 3: Calculate Reynolds Number (\(Re\))** \[ Re = \frac{\rho v d}{\mu} = \frac{950 \times 3.69 \times 0.0779}{0.001} = 272,900 \] (Turbulent flow) --- ## **Step 4: Friction Factor (\(f\))** For commercial steel pipe (assume roughness \(\epsilon = 0.046\) mm): \[ \frac{\epsilon}{d} = \frac{0.000046}{0.0779} = 0.00059 \] Using Colebrook or Swamee-Jain approximation: \[ f = 0.25 / [\log_{10}(\epsilon/3.7d + 5.74/Re^{0.9})]^2 \] \[ f = 0.25 / [\log_{10}(0.00059/3.7 + 5.74/(272900^{0.9}))]^2 \] Calculate: - \(0.00059/3.7 = 0.000159\) - \(272900^{0.9} \approx 48271\) - \(5.74/48271 = 0.000119\) - Sum: \(0.000159 + 0.000119 = 0.000278\) - \(\log_{10}(0.000278) = -3.555\) - \([ -3.555 ]^2 = 12.64\) - \(0.25 / 12.64 = 0.0198\) So **\(f \approx 0.02\)** --- ## **Step 5: Head Losses** ### **(a) Frictional Head Loss (Darcy-Weisbach)** \[ h_f = f \frac{L}{d} \frac{v^2}{2g} \] \[ \frac{L}{d} = \frac{250}{0.0779} = 3208 \] \[ \frac{v^2}{2g} = \frac{(3.69)^2}{2 \times 9.81} = \frac{13.61}{19.62} = 0.693\ \text{m} \] \[ h_{f,pipe} = 0.02 \times 3208 \times 0.693 = 44.5\ \text{m} \] ### **(b) Miscellaneous Losses** Equivalent to 500 pipe diameters: \[ L_{eq} = 500 \times 0.0779 = 38.95\ \text{m} \] \[ \frac{L}{d} = \frac{38.95}{0.0779} = 500 \] \[ h_{f,misc} = 0.02 \times 500 \times 0.693 = 6.93\ \text{m} \] ### **(c) Total Frictional Loss** \[ h_{f,total} = 44.5 + 6.93 = 51.43\ \text{m} \] --- ## **Step 6: Static Head** \[ \text{Static head (elevation)} = 6\ \text{m (column feed above pump)} - 2\ \text{m (tank above pump)} = 4\ \text{m} \] But actually, in pump calculations, the **static head** is the elevation difference between the source liquid level (2 m above pump) and the discharge point (6 m above pump): \[ \Delta z = 6\ \text{m} - 2\ \text{m} = 4\ \text{m} \] --- ## **Step 7: Pressure Head Difference** \[ \Delta P = P_{column} - P_{tank} = (200,000 - 101,300) = 98,700\ \text{Pa} \] \[ \text{Pressure head} = \frac{\Delta P}{\rho g} = \frac{98,700}{950 \times 9.81} = \frac{98,700}{9,319.5} = 10.6\ \text{m} \] --- ## **Step 8: **Total Head Required (\(H\))** \[ H = \text{Pressure head} + \text{Static head} + \text{Frictional head} \] \[ H = 10.6 + 4 + 51.43 = 66.03\ \text{m} \] --- ## **Step 9: Pump Power (Brake Power)** \[ \text{Pump Water Power} = \rho g Q H \] \[ = 950 \times 9.81 \times 0.01755 \times 66.03 = 10,788\ \text{W} = 10.79\ \text{kW} \] \[ \text{Pump Shaft Power} = \frac{\text{Pump Water Power}}{\text{Efficiency}} \] \[ = \frac{10,788}{0.7} = 15,411\ \text{W} = 15.4\ \text{kW} \] --- ### **Final Answer** **The required pump power is approximately \(15.4\) kW.** --- ### **Summary Table** | Quantity | Value | |------------------------|--------------| | Flow rate (\(Q\)) | 0.01755 m³/s | | Pipe velocity (\(v\)) | 3.69 m/s | | Reynolds number (\(Re\)) | 272,900 | | Friction factor (\(f\))| 0.02 | | Frictional head loss | 51.4 m | | Static head | 4 m | | Pressure head | 10.6 m | | **Total head (\(H\))** | **66 m** | | **Pump power** | **15.4 kW** | --- ### **Assumptions** - Pipe roughness as per commercial steel. - Sch 40 pipe internal diameter. - Minor losses as provided. - Negligible kinetic energy change at inlet/outlet. **If you need the step-by-step calculation in a spreadsheet or want to check pipe diameter, let me know!**