Problem 11 - Bernoulli's Energy Theorem

Discharge
$Q_A = Q_B = 30 \, \text{ ft}^3\text{/s}$
 

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

$\dfrac{{v_A}^2}{2g} = \dfrac{8(30^2)}{\pi^2(32.2)(18/12)^4} = 4.4752 \, \text{ ft}$

$\dfrac{{v_A}^2}{2g} = \dfrac{8(30^2)}{\pi^2(32.2)(36/12)^4} = 0.2797 \, \text{ ft}$
 

Pressure heads
$\dfrac{p_A}{\gamma_w} = \dfrac{10(12^2)}{62.4} = 23.0769 \, \text{ ft}$

$\dfrac{p_B}{\gamma_w} = \dfrac{10.9(12^2)}{62.4} = 25.1538 \, \text{ ft}$
 

Energy equation between A and B
$E_A - HL = E_B$

$\dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma_w} + z_A - HL = \dfrac{{v_B}^2}{2g} + \dfrac{p_B}{\gamma_w} + z_B$

$4.4752 + 23.0769 + 0 - HL = 0.2797 + 25.1538 + 0$

$HL = 2.1186 \, \text{ ft}$           answer