Compressible Flow credits Logo credits
Potto Home Contact Us

Potto Home

About Potto

Chapters:

  Content
  Introduction
  Sound
  Isentropic
  Shock
  Gravity
  Isothermal
  Fanno
  Rayleigh
  Tank
  Piston
  Oblique
  Prandtl-Meyer
  Hard copy
  Gas Dynamics Tables

Other things:
Other resources
Download Area
calculators

Other Resources

  FAQs
  Compare Other Books
  Articles

Potto Statistics

License

Feedback

next up previous index
Next: The Procedure for Calculating Up: Upstream Mach number, M1, Previous: Upstream Mach number, M1,   Index

The simple procedure

For example, in Figure (13.4) and (13.5), the imaginary angle is shown. The flow is far away from the object and does not ``see' the object. For example, for, $ M_1 \longrightarrow \infty$ the maximum deflection angle is calculated when $ D = Q^3 + R^2 = 0$ . This can be done by evaluating the terms $ a_1$ , $ a_2$ , and $ a_3$ for $ M_1= \infty$ .

$\displaystyle a_1 = -1 - k \sin^2\delta$    
$\displaystyle a_2 = {\left( k + 1 \right)^ 2 \sin ^2 \delta \over 4 }$    
$\displaystyle a_3 = 0$    

With these values the coefficients $ R$ and $ Q$ are
$\displaystyle R = { 9 (-) ( 1 + k \sin^2\delta ) \left({\left( k + 1 \right)^ 2 \sin ^2 \delta \over 4 } \right) - (2) (-) ( 1 + k \sin^2\delta )^2 \over 54}$    

and
$\displaystyle Q = { ( 1 + k \sin^2\delta )^2 \over 9 }$    

Figure: The view of a large inclination angle from different points in the fluid field.
\begin{figure}\centerline{\includegraphics
{cont/oblique/flowView.eps}}
\end{figure}
Solving equation (13.28) after substituting these values of $ Q$ and $ R$ provides series of roots from which only one root is possible. This root, in the case $ k=1.4$ , is just above $ \delta_{max} \sim {\pi \over 4}$ (note that the maximum is also a function of the heat ratio, $ k$ ).

While the above procedure provides the general solution for the three roots, there is simplified transformation that provides solution for the strong and and weak solution. It must be noted that in doing this transformation the first solution is ``lost'' supposedly because it is ``negative.'' In reality the first solution is not negative but rather some value between zero and the weak angle. Several researchers13.15 suggested that instead Thompson's equation should be expressed by equation (13.18) by $ \tan\theta$ and is transformed into


The solution to this equation (13.31) for the weak angle is

where these additional functions are
and

next up previous index
Next: The Procedure for Calculating Up: Upstream Mach number, M1, Previous: Upstream Mach number, M1,   Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21