Problem 01 - Bernoulli's Energy Theorem

Solve for velocity head at point (5)
$E_1 - HL_{1-5} = E_5$

$\left( \dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 \right) - HL_{1-5} = \left( \dfrac{{v_5}^2}{2g} + \dfrac{p_5}{\gamma} + z_5 \right)$

$(0 + 0 + 6) - 3 = \left( \dfrac{{v_5}^2}{2g} + 0 + 0 \right)$

$\dfrac{{v_5}^2}{2g} = 3 \, \text{ m}$
 

Velocity of flow
$v_5 = \sqrt{2g(3)} = \sqrt{2(9.81)(3)}$

$v_5 = 7.672 \, \text{ m/sec}$
 

Since the diameter of the water jet at point (5) is equal to the diameter of the pipe
$v_{pipe} = v_5$

$v_{pipe} = 7.672 \, \text{ m/sec}$           answer
 

Discharge
$Q = vA = 7.672 [ \, \frac{1}{4}\pi(0.150^2) \, ]$

$Q = 0.1356 \, \text{ m}^3\text{/sec}$

$Q = 135.6 \, \text{ Lit/sec}$          answer