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Strato limite Falkner-Skan

Le equazioni di Prandtl vengono risolte introducendo l'ipotesi che il moto del campo esterno irrotazionale possa essere espresso nella forma:

$\displaystyle \boxed{
U(x) \ =\ c\ x^m
}
$

con $ c>0$ . Si riprendano le Eq. (6.2):

\begin{displaymath}
\begin{array}{rcl}
\displaystyle \ensuremath{\frac{\partial ...
...mu \ \ensuremath{\frac{\partial^2 u}{\partial y^2}}
\end{array}\end{displaymath}

e la relazione di Bernouilli nel campo esterno allo strato limite:

$\displaystyle \ensuremath{\frac{d\,P}{d\,x}}\ = \ -\ \frac{1}{2}\ \rho \ \ensuremath{\frac{d\,U^2}{d\,x}}
$

Con il profilo della velocità esterna ipotizzato si ottiene l'espressione del gradiente di pressione in funzione della velocità esterna:

$\displaystyle \ensuremath{\frac{d\,P}{d\,x}} \ =\ -\ \rho \ c^2 \ m \ x^{2m +1}
$

sostituendo:

$\displaystyle \left( u \ensuremath{\frac{\partial u}{\partial x}} + v \ensurema...
...= c^2 \ m \ x^{2m-1}\ + \ \nu \ \ensuremath{\frac{\partial^2 u}{\partial y^2}}
$

Si procede anche ora caso come nel caso della soluzione di Blasius ricorrendo all'introduzione della funzione di corrente $ \Psi $ e alla adimensionalizzazione:

\begin{displaymath}
\begin{array}{rcl}
u \ &=&\ \displaystyle \ensuremath{\frac{...
...le -\ \ensuremath{\frac{\partial \Psi}{\partial x}}
\end{array}\end{displaymath}


$\displaystyle \eta \ $ $\displaystyle =$ $\displaystyle \ \frac{y}{\delta}$  
  $\displaystyle =$ $\displaystyle \frac{y}{\displaystyle \sqrt{\frac{\nu x}{U}}}$  
  $\displaystyle =$ $\displaystyle \displaystyle \frac{y \sqrt{c}}{\displaystyle \sqrt{\nu}}x^{\frac{m-1}{2}}$  


$\displaystyle f \ $ $\displaystyle =$ $\displaystyle \ \frac{\Psi}{\delta U}$  
  $\displaystyle =$ $\displaystyle \ \frac{\Psi}{\displaystyle U \sqrt{\frac{\nu x}{U}}}$  
  $\displaystyle =$ $\displaystyle \ \frac{\Psi}{\displaystyle \sqrt{\nu c x^{m+1}}}$  
       
$\displaystyle \Psi \ $ $\displaystyle =$ $\displaystyle \ \sqrt{\nu c} \ f\ x^{\frac{m+1}{2}}$  

Si ricavano ora tutte le espressioni delle derivate da sostituire nella equazione di Prandtl con velocità esterna di Falkner-Skan:

$\displaystyle \ensuremath{\frac{d\,\eta}{d\,x}} \ = \ \frac{y \sqrt{c}}{\sqrt{\nu}} \ \frac{m-1}{2} \ x^{\frac{m-3}{2}}
$

$\displaystyle \ensuremath{\frac{d\,\eta}{d\,y}} \ = \ \frac{\sqrt{c}}{\sqrt{\nu}} \ x^{\frac{m-1}{2}}
$


$\displaystyle \ensuremath{\frac{\partial \Psi}{\partial y}} \ $ $\displaystyle =$ $\displaystyle \ \ensuremath{\frac{\partial }{\partial y}}\left( \sqrt{\nu c} \ f \ x^{\frac{m+1}{2}} \right)$  
  $\displaystyle =$ $\displaystyle \ \sqrt{\nu c} \ x^{\frac{m+1}{2}} \ \ensuremath{\frac{d\,f}{d\,\eta}} \ \ensuremath{\frac{\partial \eta}{\partial y}}$  
  $\displaystyle =$ $\displaystyle \ \sqrt{\nu c} \ x^{\frac{m+1}{2}} \ \ensuremath{\frac{d\,f}{d\,\eta}} \ \frac{\sqrt{c}}{\sqrt{\nu}} \ x^{\frac{m-1}{2}}$  
  $\displaystyle =$ $\displaystyle c \ \ensuremath{\frac{d\,f}{d\,\eta}} \ x^m \ = \ u$  


$\displaystyle \ensuremath{\frac{\partial \Psi}{\partial x}} \ $ $\displaystyle =$ $\displaystyle \ \ensuremath{\frac{\partial }{\partial x}} \left( \sqrt{\nu c} \ f \ x^{\frac{m+1}{2}} \right)$  
  $\displaystyle =$ $\displaystyle \ \sqrt{\nu c} \ \left( x^{\frac{m+1}{2}} \ \ensuremath{\frac{d\,...
...artial \eta}{\partial x}} \ + \ f \ \frac{m+1}{2} \ x^{\frac{m+1}{2}-1} \right)$  
  $\displaystyle =$ $\displaystyle \ \sqrt{\nu c} \ \left( \ensuremath{\frac{d\,f}{d\,\eta}} \ \frac...
...-3}{2}} \ x^{\frac{m+1}{2}} \ + \ f \ \frac{m+1}{2} \ x^{\frac{m-1}{2}} \right)$  
  $\displaystyle =$ $\displaystyle \ \sqrt{\nu c} \ \left( \ensuremath{\frac{d\,f}{d\,\eta}} \ \frac...
...c{m-1}{2} \ x^{m-1} \ + \ f \ \frac{m+1}{2} \ x^{\frac{m-1}{2}} \right) \ =\ -v$  
       
    o anche elaborando ulteriormente:  
       
  $\displaystyle =$ $\displaystyle \ \frac{1}{2}\ \sqrt{\frac{\nu U}{x}} \ \left( \eta \ \ensuremath{\frac{d\,f}{d\,\eta}} \ (m-1) \ + \ f \right) \ =\ -v$  


$\displaystyle \ensuremath{\frac{\partial^2 \Psi}{\partial x \partial y }} \ $ $\displaystyle =$ $\displaystyle \ \ensuremath{\frac{\partial }{\partial x}}\left(\ensuremath{\frac{\partial \Psi}{\partial y}}\right)$  
  $\displaystyle =$ $\displaystyle \ \ensuremath{\frac{\partial }{\partial x}}\left(c \ \ensuremath{\frac{d\,f}{d\,\eta}} \ x^m\right)$  
  $\displaystyle =$ $\displaystyle \ c \ x^m \ \ensuremath{\frac{d^2\, f}{d\, \eta^2}} \ \ensuremath{\frac{d\,\eta}{d\,x}} \ +\ c \ \ensuremath{\frac{d\,f}{d\,\eta}} \ m \ x^{m-1}$  
  $\displaystyle =$ $\displaystyle \ c \ \left( \ensuremath{\frac{d\,f}{d\,\eta}} \ m \ x^{m-1} \ + ...
...2}} \ \frac{y \sqrt{c}}{\sqrt{\nu}} \ \frac{m-1}{2} \ x^{\frac{m-3}{2}} \right)$  
  $\displaystyle =$ $\displaystyle \ c \ \left( \ensuremath{\frac{d\,f}{d\,\eta}} \ m \ x^{m-1} \ + ...
...}} \ \frac{y \sqrt{c}}{\sqrt{\nu}} \ \frac{m-1}{2} \ x^{\frac{3m-3}{2}} \right)$  


$\displaystyle \ensuremath{\frac{\partial^2 \Psi}{\partial y^2}} \ $ $\displaystyle =$ $\displaystyle \ \ensuremath{\frac{\partial }{\partial y}}\left(\ensuremath{\frac{\partial \Psi}{\partial y}}\right)$  
  $\displaystyle =$ $\displaystyle \ \ensuremath{\frac{\partial }{\partial y}}\left( c \ \ensuremath{\frac{d\,f}{d\,\eta}} \ x^m \right)$  
  $\displaystyle =$ $\displaystyle \ c \ x^m \ \ensuremath{\frac{d^2\, f}{d\, \eta^2}} \ \frac{\sqrt{c}}{\sqrt{\nu}} \ x^{\frac{m-1}{2}}$  
  $\displaystyle =$ $\displaystyle \ \frac{c^{\frac{3}{2}}}{\sqrt{\nu}} \ \ensuremath{\frac{d^2\, f}{d\, \eta^2}} \ x^{\frac{3m-1}{2}}$  


$\displaystyle \frac{\partial^3\Psi}{\partial y^3} \ $ $\displaystyle =$ $\displaystyle \ \ensuremath{\frac{\partial }{\partial y}}\left( \ensuremath{\frac{\partial^2 \Psi}{\partial y^2}} \right)$  
  $\displaystyle =$ $\displaystyle \ \frac{c^{\frac{3}{2}}}{\sqrt{\nu}} \ \frac{d^3f}{d \eta^3} \ x^{\frac{3m-1}{2}} \ \frac{\sqrt{c}}{\sqrt{\nu}} \ x^{\frac{m-1}{2}}$  
  $\displaystyle =$ $\displaystyle \ \frac{c^2}{\nu} \ \frac{d^3f}{d \eta^3} \ x^{2m-1}$  

Si procede con la sostituzione:

$\displaystyle \left( u \ensuremath{\frac{\partial u}{\partial x}} + v \ensurema...
...= c^2 \ m \ x^{2m-1}\ + \ \nu \ \ensuremath{\frac{\partial^2 u}{\partial y^2}}
$

dapprima introducendo la funzione di corrente:

$\displaystyle \ensuremath{\frac{\partial \Psi}{\partial y}} \ \ensuremath{\frac...
...al y^2}} \ = \ c^2 \ m \ x^{2m-1}\ + \nu \ \frac{\partial^3\Psi}{\partial y^3}
$

e successivamente le espressioni delle derivate:

\begin{displaymath}\begin{split}c \ \ensuremath{\frac{d\,f}{d\,\eta}} \ x^m \ c ...
...rac{c^2}{\nu} \ \frac{d^3f}{d \eta^3} \ x^{2m-1} \\ \end{split}\end{displaymath}    

\begin{displaymath}\begin{split}c^2 \ x^m \ f'^2 \ m \ x^{m-1} \ + \ c^2 \ x^m \...
...\ c^2 \ m \ x^{2m-1} \ + \ c^2 \ x^{2m-1} \ f^{'''} \end{split}\end{displaymath}    

\begin{displaymath}\begin{split}f'^2 \ m \ x^{2m-1} \ + \ f' \ f'' \ \frac{m-1}{...
...}} \ = \\ = \ m \ x^{2m-1} \ + \ x^{2m-1} \ f^{'''} \end{split}\end{displaymath}    

moltiplicando tutto per $ x^{-(2m-1)}$

\begin{displaymath}\begin{split}f'^2 \ m \ + \ f' \ f'' \ \frac{m-1}{2} \ \eta \...
...\ x^{2m-1} \ x^{-(2m-1)} \ = \\ = \ m \ + \ f^{'''} \end{split}\end{displaymath}    

$\displaystyle f'^2 \ m \ - \ f \ f'' \ \frac{m+1}{2} \ = \ m \ + \ f^{'''}$    

e finalmente:

$\displaystyle \boxed{ f^{'''} \ + \ f \ f'' \ \frac{m+1}{2} \ - \ f'^2 \ m \ + \ m \ = \ 0 }$ (8.1)

   c.c.\begin{displaymath}\left\{
\begin{array}{rcll}
f(\eta=0) &=& 0 & \mbox{portata a...
...w& 1 & \mbox{$u=U$ nel campo irrotazionale}
\end{array}\right.
\end{displaymath}

che con il valore $ m=0$ esprime nuovamente la Eq. (7.2) di Blasius.

Subsections
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Next: Soluzione Up: Atti di moto laminare Previous: Atti di moto laminare   Indice
2009-01-26