$Q_A = Q_B = 0.14 \, \text{ m}^3\text{/s}$
Velocity heads
$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$
$\dfrac{{v_A}^2}{2g} = \dfrac{8(0.14^2)}{\pi^2(9.81)(0.15^4)} = 3.1990 \, \text{ m}$
$\dfrac{{v_B}^2}{2g} = \dfrac{8(0.14^2)}{\pi^2(9.81)(0.45^4)} = 0.0395 \, \text{ m}$
Pressure heads
$\dfrac{p_A}{\gamma} = \dfrac{70}{9.81} = 7.1356 \, \text{ m}$
$\dfrac{p_B}{\gamma} = \dfrac{50}{9.81} = 5.0968 \, \text{ m}$
Total head
$E = \dfrac{v^2}{2g} + \dfrac{p}{\gamma} + z$
$E_A = 3.1990 + 7.1356 + 0 = 10.3346 \, \text{ m}$
$E_B = 0.0395 + 5.0968 + 4.5 = 9.6363 \, \text{ m}$
The flow is always from higher energy to lower energy. EA > EB, thus, the flow will be from A to B. answer
Energy equation between A and B
$E_A - HL = E_B$
$10.3346 - HL = 9.6363$
$HL = 0.6983 \, \text{ m}$ answer
