Question

A pipe, having a length 200 m and 200 mm diameter with friction factor 0.015, is to be replaced by a 400 mm diameter pipe of friction factor 0.012 to convey the same quantity of flow. The equivalent length of the new pipe for the same head loss will be 
1. 8300 m
2. 8240 m
3. 8110 m
4. 8000 m

Answer 1

Correct Answer - Option 4 : 8000 m

Concept:

\({h_f} = \frac{{fL{V^2}}}{{2gD}} = \frac{{fL{Q^2}}}{{12{d^5}}}\)

∵ Q₁ = Q₂ & \({h_{{f_1}}} = {h_{{f_2}}}\)

\(\therefore \frac{{{f_1}{L_1}}}{{d_1^5}} = \frac{{{f_2}{L_2}}}{{d_2^5}}\)

Calculation:

\(\therefore \frac{{{f_1}{L_1}}}{{d_1^5}} = \frac{{{f_2}{L_2}}}{{d_2^5}}\)

\(\frac{{0.015 \times 200}}{{{{\left( {.2} \right)}^5}}} = \frac{{0.012 \times L}}{{{{\left( {.4} \right)}^5}}}\)

∴ L = 8000 m