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
Φ = ∫ A v ¯ x d A = ∫ A v ¯ r ( 2 d D ) 1 n d A = ∫ 0 D 2 v ¯ r ( 2 ( D 2 − r ) D ) 1 n ( 2 π r ) d r {\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}
Φ = 2 n + 1 n π D 1 n v ¯ r ∫ 0 D 2 ( D 2 − r ) 1 n r d r {\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}
u = D 2 − r ⇒ Φ = 2 n + 1 n π D 1 n v ¯ r ∫ 0 D 2 u 1 n ( D 2 − u ) d u {\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}
Φ = 2 n + 1 n π D 1 n v ¯ r ∫ 0 D 2 [ D 2 u 1 n − u n + 1 n ] d u {\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}
Φ = 2 n + 1 n π D 1 n v ¯ r [ D 2 n n + 1 u n + 1 n − n 2 n + 1 u 2 n + 1 n ] | 0 D 2 {\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}}}
Φ = 2 n + 1 n π D 1 n v ¯ r [ D 2 n n + 1 ( D 2 ) n + 1 n − n 2 n + 1 ( D 2 ) 2 n + 1 n ] {\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]}
Φ = 2 n + 1 n π D 1 n v ¯ r ( D 2 ) 2 n + 1 n [ n n + 1 − n 2 n + 1 ] {\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]}
Φ = π D 2 2 v ¯ r n 2 ( n + 1 ) ( 2 n + 1 ) {\displaystyle \Phi \;=\;{\frac {\pi \;D^{2}}{2}}\;{\bar {v}}_{r}\;{\frac {n^{2}}{(n\;+\;1)(2n\;+\;1)}}}
A velocidade média do escoamento é, portanto,
v ¯ = Φ A = π D 2 2 v ¯ r n 2 ( n + 1 ) ( 2 n + 1 ) ⋅ 4 π D 2 {\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}}}}
v ¯ = v ¯ r 2 n 2 ( n + 1 ) ( 2 n + 1 ) {\displaystyle {\bar {v}}\;=\;{\bar {v}}_{r}\;{\frac {2n^{2}}{(n\;+\;1)(2n\;+\;1)}}}