Problem 08 - Bernoulli's Energy Theorem

Discharge
$Q_1 = Q_2 = 0.035 \, \text{ m}^3\text{/s}$
 

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

$\dfrac{{v_1}^2}{2g} = \dfrac{8(0.035^2)}{\pi^2(9.81)(0.3^4)} = 0.0125 \, \text{ m}$

$\dfrac{{v_2}^2}{2g} = \dfrac{8(0.035^2)}{\pi^2(9.81)(0.1^4)} = 1.0122 \, \text{ m}$
 

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

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

$0.0125 + \dfrac{100}{1.03(9.81)} + 0 - HL = 1.0122 - \dfrac{15}{1.03(9.81)} + 6$

$HL = 4.38 \, \text{ m}$

$HL = 4.38 [ \, 1.03(9.81) \, ]$

$HL = 44.27 \, \text{ kPa}$           answer