# Step-by-Step Solution ## 1. **System Schematic** The system consists of: - A storage tank (at atmospheric pressure). - Centrifugal pump at the base. - 100 m long steel pipe (80 mm = .08 m ID). - Pipe fittings and valves (total frictional loss equivalent to 445 pipe diameters). - Distillation column (operating at 1.7 bar, feed point 3 m above pump suction). - Minimum tank level is 1.5 m above pump suction. **Schematic illustration (textual):** ``` [Storage Tank] (atm, 1.5 m above pump) | | (vertical drop 1.5 m) [ Pump ] | | (Pipe: 100 m + equivalent 445D) | [Column Feed] (3 m above pump, 1.7 bar) ``` *No image provided; see description above.* --- ## 2. **Known Data** - Pipe length \( L = 100 \) m - Internal diameter \( D = .08 \) m - Equivalent pipe length for fittings: \( L_{eq} = 445 D \) - Total equivalent length: \( L_{total} = L + L_{eq} \) - \( L_{eq} = 445 \times .08 = 35.6 \) m - \( L_{total} = 100 + 35.6 = 135.6 \) m Fluid properties: - Density (\( \rho \)) = 900 kg/m³ - Viscosity (\( \mu \)) = \( 1.36 \times 10^{-3} \) Pa·s Elevations: - Tank level above pump: 1.5 m - Column feed above pump: 3 m Pressures: - Tank: atmospheric (\( P_{atm} \)) - Column: 1.7 bar (\( 1.7 \times 10^5 \) N/m²) --- ## 3. **Calculate Areas and Velocities** Pipe cross-sectional area: \[ A = \frac{\pi D^2}{4} = \frac{\pi (.08)^2}{4} = .005027\, \text{m}^2 \] For each velocity \( v \): - \( Q = A \times v \) | Velocity (\( v \), m/s) | Flowrate (\( Q \), m³/s) | |-------------------------|--------------------------| | 1 | \( .005027 \) | | 2 | \( .01005 \) | | 3 | \( .01508 \) | --- ## 4. **Static Head (\( H_{static} \))** \[ H_{static} = \text{Elevation difference} + \frac{P_{column} - P_{atm}}{\rho g} \] - Elevation difference: \( 3 - 1.5 = 1.5 \) m (from tank liquid level to column feed point above pump) - \( P_{atm} = 1. \) bar = \( 1. \times 10^5 \) N/m² - \( P_{column} = 1.7 \) bar = \( 1.7 \times 10^5 \) N/m² - \( g = 9.81 \) m/s² \[ \Delta P = P_{column} - P_{atm} = (1.7 - 1.) \times 10^5 = .7 \times 10^5 = 70,000\, \text{N/m}^2 \] \[ \frac{\Delta P}{\rho g} = \frac{70,000}{900 \times 9.81} = \frac{70,000}{8,829} \approx 7.93\, \text{m} \] \[ H_{static} = 1.5 + 7.93 = 9.43\, \text{m} \] --- ## 5. **Dynamic (Friction) Head Loss (\( H_f \))** \[ H_f = f \cdot \frac{L_{total}}{D} \cdot \frac{v^2}{2g} \] Where \( f \) is the Darcy friction factor. ### **Find Reynolds Number (\( Re \)) for each \( v \):** \[ Re = \frac{\rho v D}{\mu} \] - For \( v = 1 \) m/s: \[ Re = \frac{900 \times 1 \times .08}{1.36 \times 10^{-3}} = \frac{72}{.00136} \approx 52,941 \] - For \( v = 2 \) m/s: \[ Re = \frac{900 \times 2 \times .08}{1.36 \times 10^{-3}} = \frac{144}{.00136} \approx 105,882 \] - For \( v = 3 \) m/s: \[ Re = \frac{900 \times 3 \times .08}{1.36 \times 10^{-3}} = \frac{216}{.00136} \approx 158,824 \] ### **Assume commercial steel pipe, relative roughness (\( \epsilon/D \)):** - For steel pipe, \( \epsilon \approx .046 \) mm = \( .000046 \) m - \( \epsilon/D = .000046 / .08 = .000575 \) ### **Approximate \( f \) using Swamee-Jain:** \[ f = .25 \left[ \log_{10} \left( \frac{\epsilon}{3.7D} + \frac{5.74}{Re^{.9}} \right) \right]^{-2} \] Calculate for each \( v \): #### **For \( v = 1 \) m/s:** \[ f = .25 \left[ \log_{10} \left( \frac{.000046}{3.7 \times .08} + \frac{5.74}{(52,941)^{.9}} \right) \right]^{-2} \] \[ = .25 \left[ \log_{10} \left( .000155 + .000166 \right) \right]^{-2} \] \[ = .25 \left[ \log_{10} (.000321) \right]^{-2} \] \[ \log_{10} (.000321) = -3.493 \] \[ f = .25 \left[ -3.493 \right]^{-2} = .25 \times (1/12.21) = .0205 \] #### **For \( v = 2 \) m/s:** \[ \frac{5.74}{(105,882)^{.9}} = \frac{5.74}{44,975} = .000128 \] \[ f = .25 \left[ \log_{10} (.000155 + .000128) \right]^{-2} \] \[ = .25 \left[ \log_{10}(.000283) \right]^{-2} \] \[ \log_{10}(.000283) = -3.548 \] \[ f = .25 \left[ -3.548 \right]^{-2} = .25 \times (1/12.59) = .0199 \] #### **For \( v = 3 \) m/s:** \[ \frac{5.74}{(158,824)^{.9}} = 5.74 / 63,309 = .0000907 \] \[ f = .25 \left[ \log_{10}(.000155 + .0000907) \right]^{-2} \] \[ = .25 \left[ \log_{10}(.000246) \right]^{-2} \] \[ \log_{10}(.000246) = -3.609 \] \[ f = .25 \left[ -3.609 \right]^{-2} = .25 \times (1/13.03) = .0192 \] --- ## 6. **Compute \( H_f \) for Each Velocity** \[ H_f = f \cdot \frac{L_{total}}{D} \cdot \frac{v^2}{2g} \] - \( L_{total}/D = 135.6 / .08 = 1,695 \) - \( 2g = 19.62 \) ### **For \( v = 1 \) m/s:** \[ H_f = .0205 \times 1,695 \times \frac{1^2}{19.62} = 34.75 \times .05096 = 1.77\, \text{m} \] ### **For \( v = 2 \) m/s:** \[ H_f = .0199 \times 1,695 \times \frac{4}{19.62} = 33.74 \times .2038 = 6.88\, \text{m} \] ### **For \( v = 3 \) m/s:** \[ H_f = .0192 \times 1,695 \times \frac{9}{19.62} = 32.57 \times .4588 = 14.95\, \text{m} \] --- ## 7. **Total Pump Head (\( H_{pump} \))** \[ H_{pump} = H_{static} + H_f \] | Velocity (\( v \), m/s) | \( H_{static} \) (m) | \( H_f \) (m) | \( H_{pump} \) (m) | |-------------------------|----------------------|--------------|--------------------| | 1 | 9.43 | 1.77 | 11.20 | | 2 | 9.43 | 6.88 | 16.31 | | 3 | 9.43 | 14.95 | 24.38 | --- ## 8. **Table for System Curve** | Flow velocity (\( v \), m/s) | Flow (\( Q \), m³/s) | \( H_{static} \) (m) | \( H_f \) (m) | \( H_{pump} \) (m) | |------------------------------|----------------------|----------------------|---------------|--------------------| | 1 | .005 | 9.43 | 1.77 | 11.20 | | 2 | .0101 | 9.43 | 6.88 | 16.31 | | 3 | .0151 | 9.43 | 14.95 | 24.38 | --- ## 9. **Effect of Flow Velocity on Static and Dynamic Head** - **Static head (\( H_{static} \))** does **not depend on flow velocity**; it is set by elevation and pressure difference. - **Dynamic head (\( H_f \))** increases rapidly (approximately with \( v^2 \)), since friction loss is proportional to the square of velocity. --- ## 10. **Procedure to Determine Operating Flow Rate and Pump Efficiency** 1. **Plot the System Curve:** - Plot \( H_{pump} \) vs \( Q \) from table above. 2. **Overlay Pump Curve:** - Obtain manufacturer's pump characteristic curve (head vs. flow for given speed). 3. **Find Intersection:** - The intersection of the system curve and pump curve gives the **operating point** (flow and head). 4. **Check Pump Efficiency:** - At the operating point, read the efficiency from pump's efficiency curves. 5. **Verify Power:** - Use flow, head, and efficiency to check required pump power. --- ## 11. **Required Pump Power at \( v = 2 \) m/s** ### **Calculate Flowrate** \[ Q = .01005\, \text{m}^3/\text{s} \] ### **Pump Head** \[ H_{pump} = 16.31\, \text{m} \] ### **Pump Power (hydraulic)** \[ P_{hyd} = \rho g Q H_{pump} \] \[ P_{hyd} = 900 \times 9.81 \times .01005 \times 16.31 = 900 \times 9.81 = 8,829 \] \[ 8,829 \times .01005 = 88.7 \] \[ 88.7 \times 16.31 = 1,447\, \text{W} \] ### **With Pump Efficiency (\( \eta = 75\% \))** \[ P_{pump} = \frac{P_{hyd}}{\eta} = \frac{1,447}{.75} = 1,929\, \text{W} = 1.93\, \text{kW} \] --- ##