**1 Answer**

written 2.6 years ago by | • modified 2.6 years ago |

**Assumption need to be follows :**

1) The net flow reaching any junction or node must be equal to net flow leaving the junction or node due to continuity equation.

$$∑Q_{Input}=∑Q_{Output}$$

$$∑Q_{Input}-∑Q_{Output}=0$$ Where, Q is actual flow

2) The net head loss or potential change must be zero in each loop or network.

$$∑h=0$$

Where,h is head loss

We have, Darcy - Waishbach equation

$$\begin{aligned} H_F &=\frac{4fLV^2}{2gD}\\ H_F &=\frac{4fL (\frac{4Q}{πD^2 })^2}{/2gD}\\ H_F &=\frac{32fLQ^2}{2π^2 gD^5 }\\ H_F &=(\frac{32fL}{2π^2 gD^5 } )Q^2 \\ H_F &=rQ^2 \end{aligned}$$

$$ \begin{aligned} \because Q=VA \Rightarrow V=\frac{Q}{A}=\frac{Q}{(\pi /4×D^2 )}=\frac{4Q}{\pi D^2}\end{aligned}$$

Flow in any pipe (For n = 2)

$Q_o=$ Assumed flow and

$Q=$ corrected or actual flow

$Q=Q_o+\Delta Q$

Head loss in any pipe

$H_F=rQ^2=r(Q_o+\Delta Q)^2 $

net head loss

$\sum H_F=\sum rQ^2=\sum r(Q_o+\Delta Q)^2=\sum r (Q_o^2+2Q_o\Delta Q+\Delta Q^2) $

higher powers of $\Delta Q$ can be neglected

$\sum rQ^2=\sum rQ_o^2 + \sum r \times 2Q_o \Delta Q$

For corrected network head loss should be zero

$\sum rQ^2=\sum rQ_o^2 + \sum r \times 2Q_o \Delta Q =0$

$\begin{aligned}\therefore \Delta Q=\frac{\sum rQ_o^2}{\sum 2rQ_o} \end{aligned}$

In general

$\begin{aligned}\therefore \Delta Q=-\frac{\sum rQ_o^n}{\sum nrQ_o^{n-1}} \end{aligned}$