A piping system consists of three pipes arranged in series, the l

SOLUTION

Concept:

For series connection, discharge remains same

Q_Total = Q1 = Q2 = Q3

Total head loss H_f = hf1 + hf2 + hf3

\(\begin{array}{l} f\frac{{{\rm{L}}{{\rm{V}}^2}}}{{2gD}} = {\left( {f\frac{{L{V^2}}}{{2gD}}} \right)_1} + {\left( {f\frac{{L{V^2}}}{{2gD}}} \right)_2} + {\left( {f\frac{{L{V^2}}}{{2gD}}} \right)_3}\\ f\frac{{L{Q^2}}}{{2gD.{{\left( {\frac{\pi }{4}{D^2}} \right)}^2}}} = \frac{{{f_1}{L_1}{Q^2}}}{{2g{D_1}.{{\left( {\frac{\pi }{4}D_1^2} \right)}^2}}} + \frac{{{f_2}{L_2}{Q^2}}}{{2g{D_2}{{\left( {\frac{\pi }{4}D_2^2} \right)}^2}}} \end{array}\)

Now, (∵ Q, g and V are constant)

\(∴ \frac{{fL}}{{{D^5}}} = \frac{{{f_1}{L_1}}}{{D_1^5}} + \frac{{{f_2}{L_2}}}{{D_2^5}} + \frac{{{f_3}{L_3}}}{{D_3^5}}\)

Assuming friction factor same for all, we get
\(\frac{{{L_{eq}}}}{{{{\left( {{D_{eq}}} \right)}^5}}} = \frac{{{L_1}}}{{D_1^5}} + \frac{{{L_2}}}{{D_2^5}} + \frac{{{L_3}}}{{D_3^5}}\)

Calculation:

Given:

L₁ = 1200 m, L₂ = 750 m, L₃ = 600 m, D_eq = 450 mm, D₁ = 750 mm, d₂ = 600 mm, d₃ = 450 mm

\(\frac{{{L_{eq}}}}{{{{\left( {.450} \right)}^5}}} = \frac{{1200}}{{{{\left( {.750} \right)}^5}}} + \frac{{750}}{{{{\left( {.600} \right)}^5}}} + \frac{{600}}{{{{\left( {.450} \right)}^5}}}\)