Mecânica dos fluidos/Exercícios resolvidos/E15

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Enunciado

Deduzir a expressão para o valor do coeficiente de energia cinética para um escoamento turbulento a partir da Lei da Potência.

Solução

De acordo com a definição de coeficiente de energia cinética

$\alpha =\frac{1}{{\rho }_0\mathrm{\Phi }{\overline{v}}^2}\underset{A}{\int }{\rho }_0{v}^{3}dA$

De acordo com a Lei da Potência,

$v={\overline{v}}_r{\left(\frac{2(\frac{D}{2}-r)}{D}\right)}^{\frac{1}{n}}$

Como deduzido no exercício E5, a relação entre a velocidade média e a velocidade do escoamento no centro do tubo é

$\overline{v}={\overline{v}}_r\frac{2n^2}{(n+1)(2n+1)}$

e a vazão será

$\mathrm{\Phi }=\frac{\pi D^2}{2}{\overline{v}}_r\frac{n^2}{(n+1)(2n+1)}$

Assim,

$\alpha =\frac{1}{{\rho }_0\left(\frac{\pi D^2}{2}{\overline{v}}_r\frac{n^2}{(n+1)(2n+1)}\right){\left({\overline{v}}_r\frac{2n^2}{(n+1)(2n+1)}\right)}^2}\underset{A}{\int }{\rho }_0{\left({\overline{v}}_r{\left(\frac{2(\frac{D}{2}-r)}{D}\right)}^{\frac{1}{n}}\right)}^{3}dA$

$\alpha =\frac{2^{\frac{3}{n}}(n+1{)}^3(2n+1{)}^3}{2\pi D^2{n}^6{D}^{\frac{3}{n}}}\underset{0}{\overset{\frac{D}{2}}{\int }}{(\frac{D}{2}-r)}^{\frac{3}{n}}(2\pi r)dr$

$u=\frac{D}{2}-r\Rightarrow \alpha =\frac{2^{\frac{3}{n}}(n+1{)}^3(2n+1{)}^3}{D^2{n}^6{D}^{\frac{3}{n}}}\underset{0}{\overset{\frac{D}{2}}{\int }}{u}^{\frac{3}{n}}(\frac{D}{2}-u)du$

$\alpha =\frac{2^{\frac{3}{n}}(n+1{)}^3(2n+1{)}^3}{D^2{n}^6{D}^{\frac{3}{n}}}(\frac{D}{2}\underset{0}{\overset{\frac{D}{2}}{\int }}{u}^{\frac{3}{n}}du-\underset{0}{\overset{\frac{D}{2}}{\int }}{u}^{\frac{n+3}{n}}du)$

$\alpha =\frac{2^{\frac{3}{n}}(n+1{)}^3(2n+1{)}^3}{D^2{n}^6{D}^{\frac{3}{n}}}(\frac{D}{2}\frac{n}{n+3}{\left[u^{\frac{n+3}{n}}\right]|}_0^{\frac{D}{2}}-\frac{n}{2n+3}{\left[u^{\frac{2n+3}{n}}\right]|}_0^{\frac{D}{2}})$

$\alpha =\frac{2^{\frac{3}{n}}(n+1{)}^3(2n+1{)}^3}{D^2{n}^6{D}^{\frac{3}{n}}}(\frac{D}{2}\frac{n}{n+3}{\left(\frac{D}{2}\right)}^{\frac{n+3}{n}}-\frac{n}{2n+3}{\left(\frac{D}{2}\right)}^{\frac{2n+3}{n}})$

$\alpha =\frac{2^{\frac{3}{n}}(n+1{)}^3(2n+1{)}^3}{D^2{n}^6{D}^{\frac{3}{n}}}{\left(\frac{D}{2}\right)}^{\frac{2n+3}{n}}(\frac{n}{n+3}-\frac{n}{2n+3})$

$\alpha =\frac{(n+1{)}^3(2n+1{)}^3}{4n^6}\frac{n^2}{(n+3)(2n+3)}$

$\alpha =\frac{(n+1{)}^3(2n+1{)}^3}{4n^4(n+3)(2n+3)}$

Para escoamentos plenamente turbulentos,

$\alpha =\underset{n\to \mathrm{\infty }}{\lim }\left(\frac{(n+1{)}^3(2n+1{)}^3}{4n^4(n+3)(2n+3)}\right)=1$