Para os perfis de velocidade aproximado em uma CL laminar, senoidal u=U \sin (\pi y / 2 \delta) e parabólico, u= U\left[2\left(\frac{y}{\delta}\right)-\left(\frac{y}{\delta}\right)^{2}\right], determine: \delta^{*} / \delta(\mathrm{x}) ; \theta / \delta(\mathrm{x}) e \tau_{\mathrm{w}}(\mathrm{x}) através da equação de von Karman (considere conhecidos, \rho, \mu, \mathrm{U}). Compare os resultados com a solução exata de Blasius.

Passo 1

O enunciado pede para perfis aproximados de uma camada limite laminar sobre uma placa plana com velocidade externa constante U:
– Perfil senoidal u=U\sin\bigl(\tfrac{\pi\,y}{2\,\delta}\bigr)
– Perfil parabólico u=U\bigl[2(\tfrac{y}{\delta})-(\tfrac{y}{\delta})^{2}\bigr]
Devemos obter, em função de x, as espessuras de deslocamento \delta^{*}, de quantidade de movimento \theta e a tensão de cisalhamento na parede \tau_{w}, usando a equação integral de von Kármán, que para U= const. reduz-se a
\frac{\mathrm{d}\theta}{\mathrm{d}x} =\frac{\tau_{w}}{\rho\,U^{2}}.
Também usaremos
\delta^{*} =\int_{0}^{\delta}\Bigl(1-\frac{u}{U}\Bigr)\mathrm{d}y \quad\text{e}\quad \theta =\int_{0}^{\delta}\frac{u}{U}\Bigl(1-\frac{u}{U}\Bigr)\mathrm{d}y
e
\tau_{w} =\mu\Bigl(\frac{\partial u}{\partial y}\Bigr)_{y=0}.

Passo 2

Escrevemos \eta=y/\delta e definimos f(\eta)=u/U. Então
\frac{\delta^{*}}{\delta} =\int_{0}^{1}\bigl(1-f(\eta)\bigr)\mathrm{d}\eta ,\quad \frac{\theta}{\delta} =\int_{0}^{1}f(\eta)\bigl(1-f(\eta)\bigr)\mathrm{d}\eta ,\quad \tau_{w} =\mu\,U\,\frac{f'(0)}{\delta}.
Na equação de von Kármán com \theta=\alpha\,\delta e f'(0) constante vale
\frac{\mathrm{d}\theta}{\mathrm{d}x} =\alpha\,\frac{\mathrm{d}\delta}{\mathrm{d}x} =\frac{\mu\,U\,f'(0)}{\rho\,U^{2}\,\delta} =\frac{\nu\,f'(0)}{U\,\delta},
logo
\frac{\mathrm{d}\delta}{\mathrm{d}x} =\frac{\nu\,f'(0)}{\alpha\,U\,\delta} \quad\Longrightarrow\quad \delta(x) =\sqrt{\frac{2\,\nu\,f'(0)}{\alpha\,U}\;x}\,.
E
\tau_{w}(x) =\mu\,U\,\frac{f'(0)}{\delta(x)}.

Passo 3

Perfil senoidal f(\eta)=\sin\bigl(\tfrac{\pi\,\eta}{2}\bigr).
Calculemos os coeficientes:
f'(0)=\frac{\pi}{2},
\frac{\delta^{*}}{\delta} =\int_{0}^{1}\Bigl(1-\sin\bigl(\tfrac{\pi\,\eta}{2}\bigr)\Bigr)\mathrm{d}\eta =1-\frac{2}{\pi},
\frac{\theta}{\delta} =\int_{0}^{1}\Bigl[\sin\bigl(\tfrac{\pi\,\eta}{2}\bigr) -\sin^{2}\bigl(\tfrac{\pi\,\eta}{2}\bigr)\Bigr]\mathrm{d}\eta =\frac{2}{\pi}-\frac{1}{2} =\frac{4-\pi}{2\,\pi}.
Logo \alpha=(4-\pi)/(2\pi) e f'(0)=\pi/2.

Passo 4

Substituindo em \displaystyle \delta=\sqrt{\frac{2\nu\,f'(0)}{\alpha\,U}\,x}
vem
\delta(x) =\sqrt{\frac{2\,\nu\,\dfrac{\pi}{2}}{\dfrac{4-\pi}{2\pi}\,U}\,x} =\sqrt{\frac{2\pi^{2}}{4-\pi}\,\frac{\nu\,x}{U}} =\pi\,\sqrt{\frac{2}{4-\pi}}\;\sqrt{\frac{\nu\,x}{U}}.
E
\tau_{w}(x) =\mu\,U\,\frac{f'(0)}{\delta(x)} =\mu\,U\,\frac{\tfrac{\pi}{2}}{\pi\sqrt{\tfrac{2}{4-\pi}}\sqrt{\tfrac{\nu\,x}{U}}} =\frac{\mu\,U}{2}\,\sqrt{\frac{4-\pi}{2}}\;\sqrt{\frac{U}{\nu\,x}}.

Passo 5

Perfil parabólico f(\eta)=2\eta-\eta^{2}.
Calculamos:
f'(0)=2,
\frac{\delta^{*}}{\delta} =\int_{0}^{1}\bigl(1-2\eta+\eta^{2}\bigr)\mathrm{d}\eta =\frac{1}{3},
\frac{\theta}{\delta} =\int_{0}^{1}(2\eta-\eta^{2})(1-2\eta+\eta^{2})\mathrm{d}\eta =\frac{2}{15}.
Logo \alpha=2/15.

Passo 6

Usando \displaystyle \delta=\sqrt{\frac{2\nu\,f'(0)}{\alpha\,U}\,x}
\delta(x) =\sqrt{\frac{2\,\nu\cdot2}{\tfrac{2}{15}\,U}\,x} =\sqrt{30\,\frac{\nu\,x}{U}} =\sqrt{30}\;\sqrt{\frac{\nu\,x}{U}},
e
\tau_{w}(x) =\mu\,U\,\frac{f'(0)}{\delta(x)} =\mu\,U\,\frac{2}{\sqrt{30}\sqrt{\tfrac{\nu\,x}{U}}} =\frac{2\,\mu\,U}{\sqrt{30}}\;\sqrt{\frac{U}{\nu\,x}}.

Passo 7

Comparando com a solução exata de Blasius para camada limite em placa plana ( U= const.)
\delta_{\text{Blasius}} \approx 5{,}0\;\sqrt{\frac{\nu\,x}{U}}, \quad \theta_{\text{Blasius}} \approx 0{,}664\;\sqrt{\frac{\nu\,x}{U}}, \quad \tau_{w,\text{Blasius}} =0{,}332\,\rho\,U^{2}\,\frac{1}{\sqrt{\mathrm{Re}_{x}}} =0{,}332\,\rho\,U^{2}\,\sqrt{\frac{\nu}{U\,x}}.

Resposta

Para o perfil senoidal
\frac{\delta^{*}}{\delta}=1-\frac{2}{\pi}, \quad \frac{\theta}{\delta}=\frac{4-\pi}{2\pi},
\delta(x) =\pi\sqrt{\frac{2}{4-\pi}}\;\sqrt{\frac{\nu\,x}{U}}, \quad \tau_{w}(x) =\frac{\mu\,U}{2}\sqrt{\frac{4-\pi}{2}}\;\sqrt{\frac{U}{\nu\,x}}.
Para o perfil parabólico
\frac{\delta^{*}}{\delta}=\frac{1}{3}, \quad \frac{\theta}{\delta}=\frac{2}{15},
\delta(x)=\sqrt{30}\;\sqrt{\frac{\nu\,x}{U}}, \quad \tau_{w}(x) =\frac{2\,\mu\,U}{\sqrt{30}}\;\sqrt{\frac{U}{\nu\,x}}.
Em comparação, a solução exata de Blasius dá \delta\approx5{,}0\sqrt{\nu x/U}, \theta\approx0{,}664\sqrt{\nu x/U} e \tau_{w}=0{,}332\,\rho\,U^{2}/\sqrt{\mathrm{Re}_{x}}.

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