$Q_C = Q_B = Q$
$HL_{A-C} = 1.5 \, \text{ m}$
$HL_{C-B} = 1.5 \, \text{ m}$
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
