Problem 12 - Bernoulli's Energy Theorem

Note $\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$ for circular pipes
 

Sum up energy head from A to B
$E_A - HL = E_B$

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

$(0 + 0 + 45.7) - 44.2 = \dfrac{8Q^2}{\pi^2(9.81)(0.05^4)} + 0 + 0$

$Q = 0.01065 \, \text{ m}^3\text{/s} = 10.65 \, \text{ L/s}$           answer