1. Stating the problem: Water flows into a motor through a pipe and exits through another pipe with given diameters, pressures, and a vertical height difference. We need to find the power supplied to the motor. 2. Given data: - Diameter inlet pipe $d_1=599$ mm = $0.599$ m - Diameter outlet pipe $d_2=868$ mm = $0.868$ m - Pressure inlet $P_1=14.41$ kPa = $14410$ Pa - Pressure outlet $P_2=5.50$ kPa = $5500$ Pa - Vertical height difference $z_2 - z_1 = 2.58$ m - Flow rate $Q = 532$ liters/s = $0.532$ m$^3$/s 3. Compute the cross-sectional areas: $$ A_1 = \frac{\pi}{4} d_1^2 = \frac{\pi}{4}(0.599)^2 = 0.2815 \text{ m}^2 $$ $$ A_2 = \frac{\pi}{4} d_2^2 = \frac{\pi}{4}(0.868)^2 = 0.5915 \text{ m}^2 $$ 4. Compute velocities using flow rate $Q$: $$ V_1 = \frac{Q}{A_1} = \frac{0.532}{0.2815} = 1.89 \text{ m/s} $$ $$ V_2 = \frac{Q}{A_2} = \frac{0.532}{0.5915} = 0.899 \text{ m/s} $$ 5. Use Bernoulli's equation with head loss due to work on the motor: $$ \frac{P_1}{\rho g} + z_1 + \frac{V_1^2}{2g} = \frac{P_2}{\rho g} + z_2 + \frac{V_2^2}{2g} + h_w $$ Solve for work head $h_w$: $$ h_w = \frac{P_1 - P_2}{\rho g} + z_1 - z_2 + \frac{V_1^2 - V_2^2}{2g} $$ Using $\rho=1000$ kg/m$^3$, $g=9.81$ m/s$^2$. Since $z_2 - z_1 = 2.58$ m, then $z_1 - z_2 = -2.58$ m. Calculate each term: $$ \frac{P_1 - P_2}{\rho g} = \frac{14410 - 5500}{1000 \times 9.81} = \frac{8910}{9810} = 0.9082 \text{ m} $$ $$ \frac{V_1^2 - V_2^2}{2g} = \frac{(1.89)^2 - (0.899)^2}{2 \times 9.81} = \frac{3.572 - 0.808}{19.62} = \frac{2.764}{19.62} = 0.1409 \text{ m} $$ Now sum: $$ h_w = 0.9082 - 2.58 + 0.1409 = -1.531 \text{ m} $$ The negative sign indicates power is supplied to the fluid (motor work output). 6. Power supplied $P$ is given by: $$ P = \rho g Q h_w = 1000 \times 9.81 \times 0.532 \times (-1.531) = -7987 \text{ W} $$ Magnitude is $7987$ W or $7.987$ kW. Final answer: The power supplied to the motor is approximately $7.987$ kW.