Two reservoirs connected by tow pipe lines in parallel of the same diameter D and length. It is proposed to replace the two pipe lines by a single pipeline of the same length without affecting the total discharge and loss of head due to friction. The diameter of the equivalent pipe D_e terms of the diameter of the existing pipe line, \(\frac{{{D_e}}}{D},\) is:

Concept:

When two pipes are replaced by the equivalent pipe in the case of parallel pipes then head loss across the pipes are same

\({h_f} = {h_f}_{equivalent}\)

As the head loss is equal in both cases,

\(\frac{{fL{V^2}}}{{2gD}} = \frac{{fL{V_{eq}}^2}}{{2g{D_{eq}}}}\)

And the discharge through the equivalent pipe is equal to the sum of discharge through the individual pipes

Q_eq = Q₁ + Q₂

we have, \(Q = AV = \frac{\pi }{4}{D^2}V\)

A_eq × V_eq = 2 A × V

\(\Rightarrow \frac{{\rm{\pi }}}{4}\;D_{eq}^2{V_{eq}} = 2 \times \frac{\pi }{4}{D^2}V\)

Since \({h_f}_1 = {h_f}_2 = \;{h_f}_{equivalent}\)

\(V = \sqrt {\frac{{2gD}}{{fL}}{h_f}}\)

and \({V_{eq}} = \sqrt {\frac{{2g{D_{eq}}}}{{fL}}{h_f}}\)

Now, \(\Rightarrow D_{eq}^2\sqrt {\frac{{2g{D_{eq}}}}{{fL}}{h_f}} = 2D^2\sqrt {\frac{{2gD}}{{fL}}{h_f}}\)

\(D_{eq}^{\frac{5}{2}} = 2{D^{\frac{5}{2}}}\)