• # question_answer Two identical pipes of length 'L', diameter 'd' and friction factor $f$ are connected is parallel between two points For the same total volume flow rate with pipe of same diameter 'd' and same friction factor $'f',$ the single length of the pipe will be: A) $\frac{L}{2}$                          B) $\frac{L}{\sqrt{2}}$C) $\sqrt{2}L$                  D) $\frac{L}{4}$

$Q={{Q}_{1}}+{{Q}_{2}}$ $v_{1}^{{{.}^{\centerdot }}}=\frac{4{{Q}_{1}}}{\pi d_{1}^{2}},$ ${{v}_{2}}=\frac{4{{Q}_{2}}}{\pi d_{2}^{2}}$ $\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{A}_{1}}{{v}_{1}}}{{{A}_{2}}{{v}_{2}}}=\frac{\frac{\pi }{4}d_{1}^{2}{{v}_{1}}}{\frac{\pi }{4}d_{2}^{2}{{v}_{2}}}={{\left( \frac{{{d}_{1}}}{{{d}_{2}}} \right)}^{2}}\left( \frac{{{v}_{1}}}{{{v}_{2}}} \right)$ For ${{d}_{1}}={{d}_{2}}=d$ $\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{v}_{1}}}{{{v}_{2}}}$ Frictional head loss, ${{h}_{f}}=\frac{4fl{{v}^{2}}}{2gd}$ ${{Q}_{1}}=\frac{\pi }{4}{{d}^{2}}v,$ $v=\frac{4Q}{\pi {{d}^{2}}}$ or $v\propto Q$ $\frac{4f{{L}_{e}}{{v}^{2}}}{2gd}=\frac{4f{{L}_{e}}v_{1}^{2}}{2gd}+\frac{4fLv_{2}^{2}}{2gd}$ ${{L}_{e}}{{v}^{2}}=Lv_{1}^{2}+Lv_{2}^{2}$ ${{L}_{e}}{{Q}^{2}}=LQ_{1}^{2}+LQ_{2}^{2}$ ${{Q}_{1}}={{Q}_{2}}=\frac{Q}{2}$ for same d ${{L}_{e}}{{Q}^{2}}=\frac{L{{Q}^{2}}}{4}+L\frac{{{Q}^{2}}}{4}=\frac{L{{Q}^{2}}}{2}$ ${{L}_{e}}=\frac{L}{2}$