Problem 04 - Bernoulli's Energy Theorem

Discharge
$Q_1 = Q_2 = Q = 0.015 \, \text{ m}^3\text{/s}$
 

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

$\dfrac{{v_1}^2}{2g} = \dfrac{8(0.015^2)}{\pi^2(9.81)(0.3^4)} = 0.0023 \, \text{ m}$

$\dfrac{{v_2}^2}{2g} = \dfrac{8(0.015^2)}{\pi^2(9.81)(0.1^4)} = 0.1859 \, \text{ m}$
 

 

Energy equation between 1 and 2
$E_1 - HL = E_2$

$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma_w} + z_1 - HL = \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma_w} + z_2$

$0.0023 + \dfrac{100}{9.81} + 0 - HL = 0.1859 + \dfrac{70}{9.81} + 0$

$HL = 2.874 \, \text{ m}$           answer