written 2.7 years ago by | • modified 2.7 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
$\begin{aligned}\therefore \Delta Q=-\frac{\sum rQ_o^n}{\sum nrQ_o^{n-1}} \end{aligned}$