Discharge
$Q = v_BA_B$
$Q = 3.6 \, [ \, \frac{1}{4}\pi (0.15^2) \, ]$
$Q = 0.0636 \, \text{ m}^3\text{/s}$
Output power of the pump
$P_{output} = \text{Efficiency} \times P_{input}$
$P_{output} = 0.70P_{input} = 0.70(20)$
$P_{output} = 14 \text{ hp} (746 \text{ Watts} / 1 \text{ hp})$
$P_{output} = 10~444 \, \text{ Watts}$
Head Added
$P_{output} = Q\gamma HA$
$10~444 = 0.0636(9810)HA$
$HA = 16.74 \, \text{ m}$
Velocity heads
$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$
$\dfrac{{v_A}^2}{2g} = \dfrac{8(0.0636^2)}{\pi^2(9.81)(0.2^4)} = 0.2089 \, \text{ m}$
$\dfrac{{v_B}^2}{2g} = \dfrac{8(0.0636^2)}{\pi^2(9.81)(0.15^4)} = 0.6602 \, \text{ m}$
Energy equation from A to C
$E_A + HA - HL = E_C$
$\dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma_w} + z_A + HA - HL = \dfrac{{v_C}^2}{2g} + \dfrac{p_C}{\gamma} + z_C$
$0.2089 - \dfrac{35}{9.81} + 0 + 16.74 - 3 = 0.6602 + 0 + (1.8 + h)$
$h = 7.92 \, \text{ m}$ answer
