A) \[\frac{L}{2}\]
B) \[\frac{L}{\sqrt{2}}\]
C) \[\sqrt{2}L\]
D) \[\frac{L}{4}\]
Correct Answer: A
Solution :
\[Q={{Q}_{1}}+{{Q}_{2}}\] \[v_{1}^{{{.}^{\centerdot }}}=\frac{4{{Q}_{1}}}{\pi d_{1}^{2}},\] \[{{v}_{2}}=\frac{4{{Q}_{2}}}{\pi d_{2}^{2}}\] \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{A}_{1}}{{v}_{1}}}{{{A}_{2}}{{v}_{2}}}=\frac{\frac{\pi }{4}d_{1}^{2}{{v}_{1}}}{\frac{\pi }{4}d_{2}^{2}{{v}_{2}}}={{\left( \frac{{{d}_{1}}}{{{d}_{2}}} \right)}^{2}}\left( \frac{{{v}_{1}}}{{{v}_{2}}} \right)\] 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 \[{{L}_{e}}{{Q}^{2}}=\frac{L{{Q}^{2}}}{4}+L\frac{{{Q}^{2}}}{4}=\frac{L{{Q}^{2}}}{2}\] \[{{L}_{e}}=\frac{L}{2}\]
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