Calcular o diâmetro que a tubulação do exercício E11 precisaria ter para que se garantisse uma vazão mínima de 10 l/s através de um tubo de 50 m de comprimento, considerando que a altura da água no reservatório será mantida no mínimo em 5 m.
Do exercício E12 , sabemos que
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{\displaystyle g\;\Delta h\;=\;{\frac {1}{2}}\;\left(N_{fi}\;+\;N_{fl}\;{\frac {L}{D}}\;+\;1\right){\bar {v}}^{2}\;\;\;\Rightarrow {\frac {N_{fl}}{D}}\;=\;{\frac {1}{L}}\;\left({\frac {2g\;\Delta h}{{\bar {v}}^{2}}}\;-\;N_{fi}\;-\;1\right)}
0.25
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{\displaystyle {\frac {0.25\left[log\left({\frac {e}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{-2}}{D}}\;=\;{\frac {1}{L}}\;\left({\frac {2g\;\Delta h}{\left({\frac {4\Phi }{\pi D^{2}}}\right)^{2}}}\;-\;N_{fi}\;-\;1\right)}
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L
{\displaystyle {\frac {1}{4D\cdot \left[log\left({\frac {e}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{2}}}\;=\;{\frac {\left({\frac {\pi ^{2}gD^{4}\;\Delta h}{8\Phi ^{2}}}\;-\;N_{fi}\;-\;1\right)}{L}}}
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L
{\displaystyle 4D\cdot \left[log\left({\frac {e}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{2}\cdot \left({\frac {\pi ^{2}gD^{4}\;\Delta h}{8\Phi ^{2}}}\;-\;N_{fi}\;-\;1\right)\;=\;L}
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0.15
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10
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50
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{\displaystyle 4D\cdot \left[log\left({\frac {0.15\;mm}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (10\;l/s)^{2}}}\cdot D^{4}\;-\;0,5\;-\;1\right)\;=\;50\;m}
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50
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{\displaystyle 4D\cdot \left[log\left({\frac {0.00015\;m}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot D^{4}\;-\;0,5\;-\;1\right)\;=\;50\;m}
No primeiro passo, consideremos um diâmetro de 50 mm. Assim
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1000
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0010
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50
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m
{\displaystyle N_{Re}\;=\;{\frac {4\rho _{0}\;\Phi }{\pi \mu _{0}D}}\;=\;{\frac {4\cdot 1000\;kg/m^{3}\cdot 10\;l/s}{3.14\cdot 0,0010\;kg\cdot m^{-1}\cdot s^{-1}\cdot 50\;mm}}}
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0.050
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250000
{\displaystyle N_{Re}\;=\;{\frac {4\cdot 1000\;kg/m^{3}\cdot 0.010\;m^{3}/s}{3.14\cdot 0,0010\;kg\cdot m^{-1}\cdot s^{-1}\cdot 0.050\;m}}\;=\;250000}
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{\displaystyle 4\cdot 50\;mm\cdot \left[log\left({\frac {0.00015\;m}{3.7\cdot 50\;mm}}\;+\;5.74\cdot 250000^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot (50\;mm)^{4}\;-\;0,5\;-\;1\right)\;=\;}
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{\displaystyle 4\cdot 0.050\;m\cdot \left[log\left({\frac {0.00015\;m}{3.7\cdot 0.050\;m}}\;+\;5.74\cdot 250000^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot (0.050\;m)^{4}\;-\;0,5\;-\;1\right)\;=\;-\;1,5\;m}
Consideremos um diâmetro de 300 mm. Assim
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1000
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0.30
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42000
{\displaystyle N_{Re}\;=\;{\frac {4\cdot 1000\;kg/m^{3}\cdot 0.010\;m^{3}/s}{3.14\cdot 0,0010\;kg\cdot m^{-1}\cdot s^{-1}\cdot 0.30\;m}}\;=\;42000}
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{\displaystyle 4\cdot 0.30\;m\cdot \left[log\left({\frac {0.00015\;m}{3.7\cdot 0.30\;m}}\;+\;5.74\cdot 250000^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot (0.30\;m)^{4}\;-\;0,5\;-\;1\right)\;=\;47\;m}
Consideremos um diâmetro de 310 mm. Assim
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1000
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0.31
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41000
{\displaystyle N_{Re}\;=\;{\frac {4\cdot 1000\;kg/m^{3}\cdot 0.010\;m^{3}/s}{3.14\cdot 0,0010\;kg\cdot m^{-1}\cdot s^{-1}\cdot 0.31\;m}}\;=\;41000}
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{\displaystyle 4\cdot 0.31\;m\cdot \left[log\left({\frac {0.00015\;m}{3.7\cdot 0.31\;m}}\;+\;5.74\cdot 250000^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot (0.31\;m)^{4}\;-\;0,5\;-\;1\right)\;=\;54\;m}
![{\displaystyle {\frac {0.25\left[log\left({\frac {e}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{-2}}{D}}\;=\;{\frac {1}{L}}\;\left({\frac {2g\;\Delta h}{\left({\frac {4\Phi }{\pi D^{2}}}\right)^{2}}}\;-\;N_{fi}\;-\;1\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7049c9f5a90618b8843a4a45f8ab06298036de67)
![{\displaystyle {\frac {1}{4D\cdot \left[log\left({\frac {e}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{2}}}\;=\;{\frac {\left({\frac {\pi ^{2}gD^{4}\;\Delta h}{8\Phi ^{2}}}\;-\;N_{fi}\;-\;1\right)}{L}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c77346a158b4b1e0d008ce4a23481d2f215fcac)
![{\displaystyle 4D\cdot \left[log\left({\frac {e}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{2}\cdot \left({\frac {\pi ^{2}gD^{4}\;\Delta h}{8\Phi ^{2}}}\;-\;N_{fi}\;-\;1\right)\;=\;L}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ce6f2cfce3a23c14116531987a4adb8eb6c1ad7)
![{\displaystyle 4D\cdot \left[log\left({\frac {0.15\;mm}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (10\;l/s)^{2}}}\cdot D^{4}\;-\;0,5\;-\;1\right)\;=\;50\;m}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf3526922555dbde2ed4c5710b6f912336ab92e7)
![{\displaystyle 4D\cdot \left[log\left({\frac {0.00015\;m}{3.7\;D}}\;+\;5.74\;N_{Re}^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot D^{4}\;-\;0,5\;-\;1\right)\;=\;50\;m}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7042e56f275cd3008439eeb026382be85cf7371d)
![{\displaystyle 4\cdot 50\;mm\cdot \left[log\left({\frac {0.00015\;m}{3.7\cdot 50\;mm}}\;+\;5.74\cdot 250000^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot (50\;mm)^{4}\;-\;0,5\;-\;1\right)\;=\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6570fb60bc5e1f1fa6f1569d414b49ba04bae21e)
![{\displaystyle 4\cdot 0.050\;m\cdot \left[log\left({\frac {0.00015\;m}{3.7\cdot 0.050\;m}}\;+\;5.74\cdot 250000^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot (0.050\;m)^{4}\;-\;0,5\;-\;1\right)\;=\;-\;1,5\;m}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5235657f49b93b9b98fcfde5b5359ab05b872220)
![{\displaystyle 4\cdot 0.30\;m\cdot \left[log\left({\frac {0.00015\;m}{3.7\cdot 0.30\;m}}\;+\;5.74\cdot 250000^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot (0.30\;m)^{4}\;-\;0,5\;-\;1\right)\;=\;47\;m}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6b0bc2e8433ee26272c69e93ff3af70c04c2ae1)
![{\displaystyle 4\cdot 0.31\;m\cdot \left[log\left({\frac {0.00015\;m}{3.7\cdot 0.31\;m}}\;+\;5.74\cdot 250000^{-0.9}\right)\right]^{2}\cdot \left({\frac {3.14^{2}\cdot 9.8\;m/s^{2}\cdot 5\;m}{8\cdot (0.010\;m^{3}/s)^{2}}}\cdot (0.31\;m)^{4}\;-\;0,5\;-\;1\right)\;=\;54\;m}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9577e1cd27ddb7eb0a1964c91e971bdffbf2a53)