Compressible Flow Free Textbook

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 . This can be done by evaluating the terms , , and 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

and

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

and

provides series of roots from which only one root is possible. This root, in the case

, is just above $\delta_{max} \sim {\pi \over 4}$ (note that the maximum is also a function of the heat ratio,

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 researchers^13.15 suggested that instead Thompson's equation should be expressed by equation (13.18) by $\tan\theta$ and is transformed into

$\displaystyle \left( 1 + {k-1 \over 2} {M_1}^{2} \right) \tan \delta \tan^3\the... ...) \tan^2\theta + \left( 1 + {k+1 \over 2} \right) \tan \delta \tan\theta +1 = 0$

(13.31)

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

$\displaystyle \theta_{weak} = \tan^{-1} \left( {{M_1}^2 -1 + 2 f_1(M_1,\delta) ... ...ight)} \over { 3 \left( 1 + {k-1 \over 2} {M_1}^{2} \right)\tan \delta} \right)$

(13.32)

$\displaystyle \theta_{strong} = \tan^{-1} { {{M_1}^2 -1 + 2 f_1(M_1,\delta) \co... ...ver 3\right)} \over { 3 \left( 1 + {k-1 \over 2} {M_1}^{2} \right)\tan \delta}}$

(13.33)

where these additional functions are

$\displaystyle f_1(M_1,\delta) = \sqrt{{\left({M_1}^2 -1 \right)^2 -3 \left( 1 +... ... 2} {M_1}^{2}\right) \left( 1+ {k+1 \over 2} {M_1}^{2} \right) \tan^2\delta } }$

(13.34)

and

$\displaystyle f_2(M_1,\delta) = {{ \left({M_1}^2 -1 \right)^3 - 9 \left( 1 + {k... ...} + {k+1 \over 2} {M_1}^{4} \right) \tan^2\delta } \over { f_1(M_1,\delta)^3} }$

(13.35)

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

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