Problem 13 - Bernoulli's Energy Theorem

Head lost
$HL_{1-2} = 0$

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

$HL_{3-4} = 9 \, \text{ m}$

$HL_{4-5} = 3 \, \text{ m}$
 

$HL_{1-5} = HL_{1-2} + HL_{2-3} + HL_{3-4} + HL_{4-5}$

$HL_{1-5} = 0 + 0.6 + 9 + 3$

$HL_{1-5} = 12.6 \, \text{ m}$
 

 

Energy equation between 1 and 5
$E_1 - HL_{1-5} = E_5$

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

$(0 + 0 + 30) - 12.6 = \dfrac{8Q^2}{\pi^2g{D_5}^4} + 0 + 0$

$17.4 = \dfrac{8Q^2}{\pi^2(9.81)(0.05^4)}$

$Q = 0.0363 \, \text{ m}^3\text{/s} = 36.3 \, \text{ L/s}$           answer
 

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

$\dfrac{{v_3}^2}{2g} = \dfrac{{v_4}^2}{2g} = \dfrac{8(0.0363^2)}{\pi^2(9.81)(0.15^4)} = 0.2151 \, \text{ m}$

$\dfrac{{v_5}^2}{2g} = \dfrac{8(0.0363^2)}{\pi^2(9.81)(0.05^4)} = 17.4202 \, \text{ m}$
 

Pressure heads

$\dfrac{p_2}{\gamma} = \text{El } 1 - \text{El } 2 = 30 - 25$

$\dfrac{p_2}{\gamma} = 5 \, \text{ m}$
 

Energy equation between 2 and 3
$E_2 - HL_{2-3} = E_3$

$\dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma} + z_2 - HL_{2-3} = \dfrac{{v_3}^2}{2g} + \dfrac{p_3}{\gamma} + z_3$

$(0 + 5 + 25) - 0.6 = 0.2151 + \dfrac{p_3}{\gamma} + 25$

$\dfrac{p_3}{\gamma} = 4.1849 \, \text{ m}$
 

Energy equation between 3 and 4
$E_3 - HL_{3-4} = E_4$

$\dfrac{{v_3}^2}{2g} + \dfrac{p_3}{\gamma} + z_3 - HL_{3-4} = \dfrac{{v_4}^2}{2g} + \dfrac{p_4}{\gamma} + z_4$

$(0.2151 + 4.1849 + 25) - 9 = 0.2151 + \dfrac{p_4}{\gamma} + 0$

$\dfrac{p_4}{\gamma} = 20.1849 \, \text{ m}$

 

Tabulated result

Point	Elevation head (m)	Velocity head (m)	Pressure head (m)	Total head (m)
1	30	0	0	30.0
2	25	0	5	30.0
3	25	0.2151	4.1849	29.4
4	0	0.2151	20.1849	20.4
5	0	17.4202	0	17.4