Problem 15 - Bernoulli's Energy Theorem

Discharge
$Q = v_B A_B$

$Q = 2.5 \, [ \, \frac{1}{4}\pi(0.15^2) \, ]$

$Q = 0.0442 \, \text{ m}^3\text{/s}$
 

Velocity heads
$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$

$\dfrac{{v_A}^2}{2g} = \dfrac{8(0.0442^2)}{\pi^2(9.81)(0.2^4)} = 0.1009 \, \text{ m}$

$\dfrac{{v_B}^2}{2g} = \dfrac{8(0.0442^2)}{\pi^2(9.81)(0.15^4)} = 0.3189 \, \text{ m}$
 

 

Neglecting head lost between A and B
$E_A + HA = E_B$

$\dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma} + z_A + HA = \dfrac{{v_B}^2}{2g} + \dfrac{p_B}{\gamma} + z_B$

$0.1009 - \dfrac{40}{9.81} + 0 + HA = 0.3189 + \dfrac{410}{9.81} + 2.5$

$HA = 48.5896 \, \text{ m}$
 

Power delivered by the pump
$\text{Power} = Q \gamma HA = 0.0442(9810)(48.5896)$

$\text{Power} = 21~058.37 \, \text{ Watts} \times (1 \text{ hp} / 746 \text{ Watts})$

$\text{Power} =28.23 \, \text{ hp}$           answer