Problem 18 - Bernoulli's Energy Theorem

Discharge
$Q_2 = Q_3 = Q_4 = Q$
 

Head lost
$HL_{1-2} = 1.1 \, \text{ ft}$

$HL_{2-3} = 0.7 \, \text{ ft}$

$HL_{3-4} = 2.5 \, \text{ ft}$
 

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

$\dfrac{{v_2}^2}{2g} = \dfrac{8Q^2}{\pi^2(32.2)(3/12)^4} = 6.4443Q^2$

$\dfrac{{v_3}^2}{2g} = \dfrac{{v_4}^2}{2g} = \dfrac{8Q^2}{\pi^2(32.2)(4/12)^4} = 2.039Q^2$
 

Energy equation between 1 and 4
$E_1 - HL_{1-2} - HL_{2-3} - HL_{3-4} = E_4$

$\left( \dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 \right) - HL_{1-2} - HL_{2-3} - HL_{3-4} = \left( \dfrac{{v_4}^2}{2g} + \dfrac{p_4}{\gamma} + z_4 \right)$

$(0 + 0 + 10) - 1.1 - 0.7 - 2.5 = (2.039Q^2 + 0 + 0)$

$2.039Q^2 = 5.7$

$Q = 1.672 \, \text{ ft}^3\text{/s}$           answer
 

Velocity heads at 2, 3, and 4
$\dfrac{{v_2}^2}{2g} = 6.4443(1.672^2) = 18.02 \, \text{ ft}$

$\dfrac{{v_3}^2}{2g} = \dfrac{{v_4}^2}{2g} = 2.039(1.672^2) = 5.7 \, \text{ ft}$
 

Energy equation between 1 and 2
$E_1 - HL_{1-2} = E_2$

$\left( \dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 \right) - HL_{1-2} = \left( \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma} + z_2 \right)$

$(0 + 0 + 10) - 1.1 = \left( 18.02 + \dfrac{p_2}{\gamma} + 15 \right)$

$\dfrac{p_2}{\gamma} = -24.12 \, \text{ ft}$
 

$p_2 = -24.12\gamma = -24.12(0.9 \times 62.4)$

$p_2 = -1354.58 \, \text{ psf}$

$p_2 = -1354.58 \dfrac{\text{lb}}{\text{ft}^2} \times \left( \dfrac{1 \text{ft}}{12 \text{in}} \right)^2$

$p_2 = -9.41 \, \text{ psi}$           answer
 

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

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

$\left( 5.7 + \dfrac{p_3}{\gamma} + 15 \right) - 2.5 = (5.7 + 0 + 0)$

$\dfrac{p_3}{\gamma} = -12.5 \, \text{ ft}$
 

$p_3 = -12.5\gamma = -12.5(0.9 \times 62.4)$

$p_3 = -702 \, \text{ psf}$

$p_3 = -702 \dfrac{\text{lb}}{\text{ft}^2} \times \left( \dfrac{1 \text{ft}}{12 \text{in}} \right)^2$

$p_3 = -4.875 \, \text{ psi}$           answer
 

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

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

$(18.02 - 24.12 + 15) - 0.7 = (5.7 - 12.5 + 15)$

$8.2 = 8.2$       (check!)
 

Tabulated result

Point	Elevation head (ft)	Velocity head (ft)	Pressure head (ft)	Total head (ft)
1	10	0	0	10
2	15	18.02	-24.12	8.9
3	15	5.7	-12.5	8.2
4	0	5.7	0	5.7