Problem 02 - Bernoulli's Energy Theorem

Given:
$HL_{1-2} = 0$

$HL_{2-3} = 0.6 \, \text{ m}$

$HL_{3-4} = 2.1 \, \text{ m}$

$HL_{4-5} = 0.3 \, \text{ m}$
 

Total head loss from (1) to (5)
$HL_{1-5} = HL_{1-2} + HL_{2-3} + HL_{3-4} + HL_{4-5}$

$HL_{1-5} = 0 + 0.6 + 2.1 + 0.3$

$HL_{1-5} = 3 \, \text{ m}$
 

Note:
$E = \dfrac{v^2}{2g} + \dfrac{p}{\gamma} + z$
 

Sum up head from (1) to (5)
$E_1 - HL_{1-5} = E_5$

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

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

Since the diameter of the pipe is uniform and the opening for the jet is equal to the diameter of the pipe, the velocity heads at any point on the pipe are equal. Thus,

$\dfrac{{v_3}^2}{2g} = \dfrac{{v_4}^2}{2g} = \dfrac{{v_5}^2}{2g} = 3 \, \text{ m}$
 

 

Sum up head from (1) to (2)
$E_1 - HL_{1-2} = E_2$

$(0 + 0 + 6) - 0 = \left( 0 + \dfrac{p_2}{\gamma} + 0 \right)$

$\dfrac{p_2}{\gamma} = 6 \, \text{ m}$
 

Sum up head from (2) to (3)
$E_2 - HL_{2-3} = E_3$

$(0 + 6 + 0) - 0.6 = \left( 3 + \dfrac{p_3}{\gamma} + 0 \right)$

$\dfrac{p_3}{\gamma} = 2.4 \, \text{ m}$
 

Sum up head from (3) to (4)
$E_3 - HL_{3-4} = E_4$

$(3 + 2.4 + 0) - 2.1 = \left( 3 + \dfrac{p_4}{\gamma} + 0 \right)$

$\dfrac{p_4}{\gamma} = 0.3 \, \text{ m}$
 

Sum up head from (4) to (5)
$E_4 - HL_{4-5} = E_5$

$(3 + 0.3 + 0) - 0.3 = (3 + 0 + 0)$

$3 = 3 \, \text{(check)}$
 

Tabulated result

Point	Elevation head (m)	Velocity head (m)	Pressure head (m)	Total head (m)
1	6	0	0	6.0
2	0	0	6	6.0
3	0	3	2.4	5.4
4	0	3	0.3	3.3
5	0	3	0	3.0

 

Piezometric heights
$h = \dfrac{p}{\gamma} + z$

$h_3 = 0 + 2.4 = 2.4 \, \text{ m}$           answer

$h_4 = 0 + 0.3 = 0.3 \, \text{ m}$           answer