$Q_C = Q_B = Q$

$HL_{A-C} = 1.5 \, \text{ m}$

$HL_{C-B} = 1.5 \, \text{ m}$

Velocity head at B and C in terms of Q

$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4} = \dfrac{8Q^2}{\pi^2(9.81)(0.15^4)}$

$\dfrac{v^2}{2g} = 163.21Q^2$

Energy Equation between A and B

$E_A - HL_{A-C} - HL_{C-B} = E_B$

$(0 + 0 + 0) - 1.5 - 1.5 = 163.21Q^2 + 0 - 4.2$

$163.21Q^2 = 1.2$

$Q = 0.0857 \, \text{ m}^3\text{/s}$ *answer*

$Q = 85.7 \dfrac{\text{Liters}}{\text{sec}} \times \dfrac{1 \text{gallon}}{3.78 \text{Liters}} \times \dfrac{60 \text{sec}}{1 \text{min}}$

$Q = 1360.32 \, \text{ gallons/min}$ *answer*

Thus, the velocity head at B and C is

$\dfrac{v^2}{2g} = 163.21(0.0857^2) = 1.2 \, \text{ m}$

Energy equation between A and C

$E_A - HL_{A-C} = E_C$

$(0 + 0 + 0) - 1.5 = 1.2 + \dfrac{p_C}{\gamma} - 1.8$

$\dfrac{p_C}{\gamma} = -0.9 \, \text{ m}$

$p_C = 1.5\gamma = -0.9(9.81)$

$p_C = -8.829 \, \text{ kPa}$ *answer*

## Comments

## Sir pa check po yung

Sir pa check po yung elevation head ng point C sa last part ng solution..ty..why po -4.2?...dapat 1.8 po dba?

## Salamat.

Salamat. Na koreksyunan na.