Mecânica dos fluidos/Exercícios resolvidos/E5: diferenças entre revisões

<center><math>\Phi \;=\; \int_A \bar vv_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 </math></center>

<center><math>\Phi \;=\; \frac{42 ^ {\frac{n \;+\; 1}{n}} \; \pi}{D^{\frac{1}{n}}} \; \bar v_r \; \int_0^{\frac{D}{2}} \; \left[ \left( \frac{D}{2} \;-\; r \right) ^ {\frac{1}{n}} r \;-\; r ^{\frac{n \;+\; 1}{n}} \right] dr </math></center>

<center><math>\Phiu \;=\; \frac{4 \piD}{D2} \; -\bar; v_rr \;\;\; \left.Rightarrow \left[Phi \left(;=\; \frac{D}{2} \right) ^ {\frac{n \;+\; 1}{n}} \frac; \pi}{rD^2}{2} \;-\; \frac{n1}{2nn}}} \;+ \;bar 1}v_r \; r \int_0^{\frac{2nD}{2}} \;+\; u^{\frac{1}{n}} \right]; \right|_0^{left( \frac{D}{2}} \;-\; u \right) \; du</math></center>

<center><math>\Phi \;=\; \frac{42 ^ {\pi}frac{D}n \; \bar v_r +\; \left[ \left( \frac{D1}{2n}} \right); \pi}{D^ {\frac{1}{n}}} \; \bar v_r \; \int_0^{\frac{D^2}{82}} \;-\;left[ \frac{nD}{2n \;+\; 12} \; \left( u^{\frac{D1}{2n}} \right);-\; u^{\frac{2nn \;+\; 1}{n}} \right] \; du</math></center>

<center><math>\Phi \;=\; \frac{42 ^ {\pifrac{n \;+\; 1}{Dn}} \; \bar v_r pi}{D ^ {\frac{2n \;+\; 1}{n}}} \; \bar v_r \; \left[. \left([ \frac{2^nD}{82} \;-\; \frac{n}{2nn \;+\; 1} \; u^{\left(frac{n \;+\; 1}{n}} \;-\; \frac{Dn}{22n \;+\; 1} \right); u^{\frac{2n \;+\; 1}{n}} \right] \right|_0^{\frac{D}{2}} </math></center>