Velocity heads in terms of discharge Q
$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$
$\dfrac{{v_1}^2}{2g} = \dfrac{8Q^2}{\pi^2(9.81)(0.075^4)} = 2611.42Q^2$
$\dfrac{{v_2}^2}{2g} = \dfrac{8Q^2}{\pi^2(9.81)(0.025^4)} = 211~524.75Q^2$
Head lost
$HL = 0.05\left( \dfrac{{v_2}^2}{2g} \right) = 0.05(211~524.75Q^2)$
$HL = 10576.24Q^2$
Energy equation between 1 and 2
$E_1 - HL = E_2$
$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma_w} + z_1 - HL = \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma_w} + z_2$
$2611.42Q^2 + 30 + 0 - 10576.24Q^2 = 211~524.75Q^2 + 0 + 0$
$219~489.57Q^2 = 30$
$Q = 0.01169 \, \text{ m}^3\text{/s}$
Velocity head at point 2
$\dfrac{{v_2}^2}{2g} = 211~524.75(0.01169^2)$
$\dfrac{{v_2}^2}{2g} = 28.91 \, \text{ m}$ answer
