Deduzir a relação entre a velocidade média do escoamento turbulento e o valor da velocidade no centro de um duto de seção circular, usando a lei de potências. Assumir que a lei vale inclusive para os pontos bastante próximos às paredes do tubo.
A vazão no duto será
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{\displaystyle \Phi \;=\;\int _{A}{\bar {v}}_{x}\;dA\;=\;\int _{A}{\bar {v}}_{r}\;\left({\frac {2d}{D}}\right)^{\frac {1}{n}}\;dA\;=\;\int _{0}^{\frac {D}{2}}{\bar {v}}_{r}\;\left({\frac {2({\frac {D}{2}}\;-\;r)}{D}}\right)^{\frac {1}{n}}\;(2\pi r)dr}
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{\displaystyle \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\int _{0}^{\frac {D}{2}}\;\left({\frac {D}{2}}\;-\;r\right)^{\frac {1}{n}}\;r\;dr}
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{\displaystyle u\;=\;{\frac {D}{2}}\;-\;r\;\;\;\Rightarrow \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\int _{0}^{\frac {D}{2}}\;u^{\frac {1}{n}}\;\left({\frac {D}{2}}\;-\;u\right)\;du}
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{\displaystyle \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\int _{0}^{\frac {D}{2}}\left[{\frac {D}{2}}\;u^{\frac {1}{n}}\;-\;u^{\frac {n\;+\;1}{n}}\right]\;du}
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{\displaystyle \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\left.\left[{\frac {D}{2}}\;{\frac {n}{n\;+\;1}}\;u^{\frac {n\;+\;1}{n}}\;-\;{\frac {n}{2n\;+\;1}}\;u^{\frac {2n\;+\;1}{n}}\right]\right|_{0}^{\frac {D}{2}}}
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{\displaystyle \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\left[{\frac {D}{2}}\;{\frac {n}{n\;+\;1}}\left({\frac {D}{2}}\right)^{\frac {n\;+\;1}{n}}\;-\;{\frac {n}{2n\;+\;1}}\left({\frac {D}{2}}\right)^{\frac {2n\;+\;1}{n}}\right]}
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{\displaystyle \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\left({\frac {D}{2}}\right)^{\frac {2n\;+\;1}{n}}\left[{\frac {n}{n\;+\;1}}\;-\;{\frac {n}{2n\;+\;1}}\right]}
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{\displaystyle \Phi \;=\;{\frac {\pi \;D^{2}}{2}}\;{\bar {v}}_{r}\;{\frac {n^{2}}{(n\;+\;1)(2n\;+\;1)}}}
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{\displaystyle {\bar {v}}\;=\;{\frac {\Phi }{A}}\;=\;{\frac {\pi \;D^{2}}{2}}\;{\bar {v}}_{r}\;{\frac {n^{2}}{(n\;+\;1)(2n\;+\;1)}}\cdot {\frac {4}{\pi \;D^{2}}}}
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{\displaystyle {\bar {v}}\;=\;{\bar {v}}_{r}\;{\frac {2n^{2}}{(n\;+\;1)(2n\;+\;1)}}}
![{\displaystyle \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\int _{0}^{\frac {D}{2}}\left[{\frac {D}{2}}\;u^{\frac {1}{n}}\;-\;u^{\frac {n\;+\;1}{n}}\right]\;du}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b16befe8ee25cacfd5b5f8c57a83cf166b51fa1)
![{\displaystyle \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\left.\left[{\frac {D}{2}}\;{\frac {n}{n\;+\;1}}\;u^{\frac {n\;+\;1}{n}}\;-\;{\frac {n}{2n\;+\;1}}\;u^{\frac {2n\;+\;1}{n}}\right]\right|_{0}^{\frac {D}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/630df1c50d0338de584d86f9e0f4637fb6148e7d)
![{\displaystyle \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\left[{\frac {D}{2}}\;{\frac {n}{n\;+\;1}}\left({\frac {D}{2}}\right)^{\frac {n\;+\;1}{n}}\;-\;{\frac {n}{2n\;+\;1}}\left({\frac {D}{2}}\right)^{\frac {2n\;+\;1}{n}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27c4f04dac51aaad16c81bf4d2188f0733deed50)
![{\displaystyle \Phi \;=\;{\frac {2^{\frac {n\;+\;1}{n}}\;\pi }{D^{\frac {1}{n}}}}\;{\bar {v}}_{r}\;\left({\frac {D}{2}}\right)^{\frac {2n\;+\;1}{n}}\left[{\frac {n}{n\;+\;1}}\;-\;{\frac {n}{2n\;+\;1}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3f76fa74e4d195bbf95fc1360ddcb3b302e83a5)
