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Momentum Integral
We can write equation (4.77) in the form
![\begin{displaymath}-\nu\frac{\partial^{2}u}{\partial y^{2}} = \frac{\partial}{\p...
...u\frac{\partial u}{\partial x}-v\frac{\partial u}{\partial y},
\end{displaymath}](img119.gif) |
(89) |
and if we multiply equation (4.78) by (u-U) we obtain
![\begin{displaymath}(U-u)\frac{\partial u}{\partial x}+(U-u)\frac{\partial v}{\partial y} = 0.
\end{displaymath}](img120.gif) |
(90) |
If we add these two equations, we obtain
![\begin{displaymath}-\nu\frac{\partial^{2}u}{\partial y^{2}} = \frac{\partial}{\p...
...ac{\partial U}{\partial x}+\frac{\partial}{\partial x}(vU-vu).
\end{displaymath}](img121.gif) |
(91) |
Next, we integrate with respect to y from 0 to
.
This yields
![\begin{displaymath}-\nu\frac{\partial u}{\partial y}\vert _{y=0} = \frac{\partia...
...rtial U}{\partial x}\int_{0}^{\infty}(U-u)dy+(vU)\vert _{y=0},
\end{displaymath}](img123.gif) |
(92) |
since
and v(U-u) tend to zero as
.
If there is no suction at the body surface,
vy=0 = 0. We assume that there is no suction at the body surface, and by taking into account equations (4.79) to (4.82), we can write equation (5.92) in the form
![\begin{displaymath}\frac{\tau_{\omega}}{\rho} = \nu\frac{\partial u}{\partial y}...
...l x}(U^{2}\delta_{2})+U\frac{\partial U}{\partial x}\delta_{1}
\end{displaymath}](img125.gif) |
(93) |
Next: Energy Integral.
Up: Momentum and Energy Equations.
Previous: Momentum and Energy Equations.
Karl P Burr
2003-03-12