1. **Problem statement:** A 21-hp pump operates at 75% efficiency, drawing water from a 216 mm diameter suction pipe and discharging through a 148 mm diameter pipe where the velocity is 3.57 m/s. The pressure at point A in the suction pipe is 35.94 kPa below atmospheric pressure. Point A is 1.56 m below point B on the 148 mm line. We need to find the maximum elevation above B to which water can be raised, considering a friction head loss of 2.91 m. 2. **Convert pump horsepower to power input:** $$\text{Power input} = 21 \times 746 = 15666\text{ W}$$ 3. **Calculate power output using efficiency:** $$\text{Power output} = 0.75 \times 15666 = 11749.5\text{ W}$$ 4. **Calculate cross-sectional areas of pipes:** $$A_1 = \pi \left(\frac{0.216}{2}\right)^2 = 0.03663\, m^2$$ $$A_2 = \pi \left(\frac{0.148}{2}\right)^2 = 0.01718\, m^2$$ 5. **Calculate velocity in the suction pipe (using continuity equation):** $$Q = A_2 v_2 = 0.01718 \times 3.57 = 0.06134\, m^3/s$$ $$v_1 = \frac{Q}{A_1} = \frac{0.06134}{0.03663} = 1.674\, m/s$$ 6. **Calculate pressure at point A (given as below atmospheric):** $$P_A = -35.94 \times 10^3\, Pa$$ 7. **Use Bernoulli equation between points A and B including pressure, velocity, elevation and head loss:** $$\frac{P_A}{\rho g} + \frac{v_1^2}{2g} + z_A = \frac{P_B}{\rho g} + \frac{v_2^2}{2g} + z_B + h_f + h_p$$ Assuming point B discharges to atmosphere: $$P_B = 0$$ And $$z_A = -1.56\, m$$ 8. **Calculate specific values:** Density of water $$\rho = 1000\, kg/m^3$$ Acceleration due to gravity $$g = 9.81\, m/s^2$$ Plug in values: $$\frac{-35940}{1000 \times 9.81} + \frac{1.674^2}{2 \times 9.81} -1.56 = 0 + \frac{3.57^2}{2 \times 9.81} + 0 + 2.91 + h_p$$ Calculate each term: $$\frac{-35940}{9810} = -3.662\, m$$ $$\frac{1.674^2}{19.62} = 0.143\, m$$ $$\frac{3.57^2}{19.62} = 0.650\, m$$ Equation becomes: $$-3.662 + 0.143 - 1.56 = 0 + 0.650 + 2.91 + h_p$$ $$-5.079 = 3.56 + h_p$$ Solve for $$h_p$$ (pump head): $$h_p = -5.079 - 3.56 = -8.639\, m$$ This negative value indicates energy is added; for pump head, take magnitude: $$h_p = 8.639\, m$$ 9. **Calculate total head produced by pump related to power output:** $$h_p = \frac{P}{\rho g Q} = \frac{11749.5}{1000 \times 9.81 \times 0.06134} = 19.54\, m$$ The negative sign above is from rearrangement; actual head provided by pump is 19.54 m. 10. **Calculate maximum elevation rise $$z$$ above B using energy head:** The energy head is available for elevation rise after overcoming velocity head and losses: $$z = h_p + \frac{P_A}{\rho g} + \frac{v_1^2}{2g} - \frac{v_2^2}{2g} - h_f - z_A$$ Plug in values: $$z = 19.54 - 3.662 + 0.143 - 0.650 - 2.91 + 1.56 = 14.022\, m$$ **Final answer:** The maximum elevation above point B to which the water can be raised is **14.022 meters**.