# Centrifugal Pump Impeller Calculations Given data: - **Pump speed, \( N \):** 125 rpm - **Flow rate, \( Q \):** 150 gpm (gallons per minute) - **Inlet radius, \( r_1 \):** 3 in - **Outlet radius, \( r_2 \):** 9.75 in - **Inlet blade width, \( b_1 \):** 1.5 in - **Outlet blade width, \( b_2 \):** 1.125 in - **Inlet blade angle, \( \beta_1 \):** 60° - **Outlet blade angle, \( \beta_2 \):** 70° Let's solve each part step by step. --- ## Step 1: Unit Conversions ### Flow rate \((Q)\) \[ Q = 150 \text{ gpm} \times .13368 = 200.52 \text{ ft}^3/\text{min} = \frac{200.52}{60} = 3.342 \text{ ft}^3/\text{s} \] ### Radii and widths (inches to feet) \[ r_1 = \frac{3}{12} = .25 \text{ ft} \] \[ r_2 = \frac{9.75}{12} = .8125 \text{ ft} \] \[ b_1 = \frac{1.5}{12} = .125 \text{ ft} \] \[ b_2 = \frac{1.125}{12} = .09375 \text{ ft} \] --- ## Step 2: Velocities ### Angular velocity (\(\omega\)): \[ \omega = 2\pi N / 60 = 2\pi \times 125 / 60 = 130.90 \text{ rad/min} = 2.182 \text{ rad/s} \] ### Peripheral (tangential) velocity: \[ u_1 = r_1 \omega = .25 \times 2.182 = .546 \text{ ft/s} \] \[ u_2 = r_2 \omega = .8125 \times 2.182 = 1.773 \text{ ft/s} \] --- ## Step 3: Flow Velocity at Outlet (\(V_{r2}\)) \[ A_2 = 2\pi r_2 b_2 = 2\pi \times .8125 \times .09375 = .479 \text{ ft}^2 \] \[ V_{r2} = Q / A_2 = 3.342 / .479 = 6.98 \text{ ft/s} \] --- ## Step 4: Theoretical Head (\(H_p\)) Theoretical head delivered by the pump: \[ H_p = \frac{u_2 V_{u2} - u_1 V_{u1}}{g} \] Where: - \( V_{u2} = u_2 - V_{r2}\cot\beta_2 \) - \( V_{u1} = u_1 - V_{r1}\cot\beta_1 \) \[ \cot 70^\circ = .364, \quad \cot 60^\circ = .577 \] \[ V_{u2} = u_2 - V_{r2} \cot 70^\circ = 1.773 - 6.98 \times .364 = 1.773 - 2.54 = -.767 \text{ ft/s} \] Now, let's find \( V_{r1} \): \[ A_1 = 2\pi r_1 b_1 = 2\pi \times .25 \times .125 = .196 \text{ ft}^2 \] \[ V_{r1} = Q / A_1 = 3.342 / .196 = 17.06 \text{ ft/s} \] \[ V_{u1} = u_1 - V_{r1} \cot 60^\circ = .546 - 17.06 \times .577 = .546 - 9.84 = -9.29 \text{ ft/s} \] \[ u_2 V_{u2} - u_1 V_{u1} = (1.773 \times -.767) - (.546 \times -9.29) = -1.36 + 5.08 = 3.72 \] \[ g = 32.174 \text{ ft/s}^2 \] \[ H_p = \frac{3.72}{32.174} = .1156 \text{ ft} \] --- ## Step 5: Mechanical Power (\( \dot{W}_m \)) \[ \dot{W}_m = \rho g Q H_p \] \[ \rho = 62.4 \text{ lb/ft}^3 \] \[ \dot{W}_m = 62.4 \times 32.174 \times 3.342 \times .1156 = 778 \text{ ft-lb/s} \] \[ 1 \text{ hp} = 550 \text{ ft-lb/s} \] \[ \dot{W}_m = 778 / 550 = 1.41 \text{ hp} \] --- ## **Summary of Answers** ### (a) Mechanical Power \( \dot{W}_m \) \[ \boxed{1.41 \text{ hp}} \] ### (b) Theoretical Head \( H_p \) \[ \boxed{.116 \text{ ft}} \] --- ### **Notes** - Double-check flow rate and velocity conversions for your specific units and significant digits required by your instructor. - If you need the answer in SI units or more detailed breakdown, let me know!