$h_f = \dfrac{0.0826fLQ^2}{D^5}$
$h_{f1} = \dfrac{0.0826(0.018)(5000)(0.045^2)}{0.3^5} = 6.195 ~ \text{m}$
$\text{Elev of } P’ = \text{Elev of } A - h_{f1} = 80 - 6.195$
$\text{Elev of } P’ = 73.805 ~ \text{m}$ ← [ B ] answer for part 1
The energy at P, denoted as P', is higher than reservoir B, thus, the flow in pipe 2 is towards B
$h_{f2} = \text{Elev of }P' - \text{Elev of } B = 73.805 - 70$
$h_{f2} = 3.805 ~ \text{m}$
$\dfrac{0.0826(0.018)(4000){Q_2}^2}{0.25^5} = 3.085$
$Q_2 = 0.025 ~ \text{m}^3/\text{s}$ ← [ A ] answer for part 2
At junction P, inflow = outflow:
$Q_1 = Q_2 + Q_3$
$0.045 = 0.025 + Q_3$
$Q_3 = 0.02 ~ \text{m}^3/\text{s}$
$h_{f3} = \text{Elev of } P' - \text{Elev of } C = 73.805 - 60$
$h_{f3} = 13.805 ~ \text{m}$
$\dfrac{0.0826(0.018)(3500)(0.02^2)}{{D_3}^5} = 13.805$
$D_3 = 0.172 ~ \text{m} = 172 ~ \text{mm}$ ← [ D ] answer for part 3
