# 12.2.2.1: Large deflection angle for given, $$M_1$$

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The first range is when the deflection angle reaches above the maximum point. For a given upstream Mach number, $$M_1$$, a change in the inclination angle requires a larger energy to change the flow direction. Once, the inclination angle reaches the "maximum potential energy,'' a change in the flow direction is no longer possible. As the alternative view, the fluid "sees'' the disturbance (in this case, the wedge) in front of it and hence the normal shock occurs. Only when the fluid is away from the object (smaller angle) liquid "sees'' the object in a different inclination angle. This different inclination angle is sometimes referred to as an imaginary angle.

## The Simple Calculation Procedure

For example, in Figure 12.4 and ??, 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 $$M_1 = \infty$$.
\begin{align*}
a_1 & = -1 - k \sin^2\delta \\
a_2 & = \dfrac{\left( k + 1 \right)^ 2 \sin ^2 \delta }{ 4 } \\
a_3 & = 0
\end{align*}
With these values the coefficients $$R$$ and $$Q$$ are
\begin{align*}
R = \dfrac{ - 9 ( 1 + k \sin^2\delta )
\left(\dfrac{\left( k + 1 \right)^ 2 \,
\sin ^2 \delta }{ 4 } \right)
- (2) (-) ( 1 + k\, \sin^2\delta )^2 }{ 54}
\end{align*}
and
\begin{align*}
Q = \dfrac{ ( 1 + k\, \sin^2\delta )^2 }{ 9 }
\end{align*}

Fig. 12.5 The view of a large inclination angle from different

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

$\left( 1 + \dfrac{k-1 }{ 2} {M_1}^{2} \right) \tan \delta \tan^3\theta - \left({M_1}^{2} - 1 \right) \tan^2\theta + \left( 1 + {k+1 \over 2} \right) \tan \delta \tan\theta +1 = 0 \label{2Dgd:eq:Oemanuel}$

The solution to this equation (31) for the weak angle is

Weak Angle Solution

$\label{2Dgd:eq:OweakThompson} \theta_{weak} = \tan^{-1} \left( \dfrac{ {{M_1}^2 -1 + 2 \,f_1(M_1,\delta) \, \cos \left( \dfrac{4\,\pi + \cos^{-1}(f_2(M_1,\delta)) }{ 3} \right)} }{ { 3 \,\left( 1 + \dfrac{k-1 }{ 2} \, {M_1}^{2} \right)\,\tan \delta} } \right)$

Strong Angle Solution

$\label{2Dgd:eq:OstrongThompson} \theta_{strong} = \tan^{-1} \, \dfrac{ ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Civil_Engineering/Book:_Fluid_Mechanics_(Bar-Meir)/12:_Compressible_Flow_2–Dimensional/12.2:_Oblique_Shock/12.2.2:_When_No_Oblique_Shock_Exist_or_the_case_of_$$D%3E0$$/12.2.2.1:_Large_deflection_angle_for_given,_$$M_1$$), /content/body/div[1]/div[2]/p[3]/span, line 1, column 4 $

$f_1(M_1,\delta) = \sqrt{{\left({M_1}^2 -1 \right)^2 -3 \, \left( 1 + \dfrac{k-1}{ 2} {M_1}^{2}\right) \left( 1+ \dfrac{k+1 }{ 2} {M_1}^{2} \right) \tan^2\delta } } \label{2Dgd:eq:Ofmtheta1}$

and

$f_2(M_1,\delta) = \dfrac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Civil_Engineering/Book:_Fluid_Mechanics_(Bar-Meir)/12:_Compressible_Flow_2–Dimensional/12.2:_Oblique_Shock/12.2.2:_When_No_Oblique_Shock_Exist_or_the_case_of_$$D%3E0$$/12.2.2.1:_Large_deflection_angle_for_given,_$$M_1$$), /content/body/div[1]/p[8]/span, line 1, column 2  ^2\). Once $$M_{1n}$$ is found, then the Mach angle can be easily calculated by equation (8). To compare these two equations the simple case of Maximum for an infinite Mach number is examined. It must be pointed out that similar procedures can also be proposed (even though it does not appear in the literature). Instead, taking the derivative with respect to $$\theta$$, a derivative can be taken with respect to $$M_1$$. Thus, \[ \dfrac{d \tan \delta }{ dM_1} = 0 \label{2Dgd:eq:OmaxMa}$ and then solving equation (39) provides a solution for $$M_{max}$$. A simplified case of the Maximum Deflection Mach Number's equation for large Mach number becomes

${M_{1n}} = \sqrt{\dfrac{ k+1 }{ 2\,k } } M_{1} \quad \text{for} \quad M_{1} >> 1 \label{2Dgd:eq:OmenikoffLarge}$ Hence, for large Mach numbers, the Mach angle is $$\sin\theta = \sqrt{ k+1\over 2k }$$ (for k=1.4), which makes $$\theta = 1.18$$ or $$\theta = 67.79^{\circ}$$. With the value of $$\theta$$ utilizing equation (12), the maximum deflection angle can be computed. Note that this procedure does not require an approximation of $$M_{1n}$$ to be made. The general solution of equation (36) is

Normal Shock Minikoff Solution

$\label {2Dgd:eq:OminikoffSol} M_{1n} = \dfrac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Civil_Engineering/Book:_Fluid_Mechanics_(Bar-Meir)/12:_Compressible_Flow_2–Dimensional/12.2:_Oblique_Shock/12.2.2:_When_No_Oblique_Shock_Exist_or_the_case_of_$$D%3E0$$/12.2.2.1:_Large_deflection_angle_for_given,_$$M_1$$), /content/body/div[1]/div[3]/p[3]/span[1], line 1, column 1  ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Civil_Engineering/Book:_Fluid_Mechanics_(Bar-Meir)/12:_Compressible_Flow_2–Dimensional/12.2:_Oblique_Shock/12.2.2:_When_No_Oblique_Shock_Exist_or_the_case_of_$$D%3E0$$/12.2.2.1:_Large_deflection_angle_for_given,_$$M_1$$), /content/body/div[1]/div[3]/p[3]/span[2], line 1, column 2 $

Note that Maximum Deflection Mach Number's equation can be extended to deal with more complicated equations of state (aside from the perfect gas model). This typical example is for those who like mathematics.

Example 12.1

Derive the perturbation of Maximum Deflection Mach Number's equation for the case of a very small upstream Mach number number of the form $$M_1 = 1 + \epsilon$$. Hint, Start with equation (36) and neglect all the terms that are relatively small.

Solution 12.1

The solution can be done by substituting ($$M_1 = 1 + \epsilon$$) into equation (36) and it results in

Normal Shock Small Values

$\label {2Dgd:eq:OsmallMachMaxDeflction} M_{1n}= {\sqrt{\dfrac{\sqrt{\epsilon (k) } + {\epsilon^2} +2\,\epsilon - 3 + k\, \epsilon^2+2\,k\epsilon+k }{ 4\,k} } }$

where the epsilon function is

$\epsilon(k) = (k^2+2k+1 )\,\epsilon^4+(4\,k^2+8\,k+4)\,\epsilon^3 + \ (14\,k^2+12\,k - 2)\,\epsilon^2+( 20\,k^2+8\,k-12) \,\epsilon + 9\,\left(k+1\right)^2 \label{2Dgd:eq:OepsilonF}$

Now neglecting all the terms with $$\epsilon$$ results for the epsilon function in

$\epsilon(k) \sim 9\,\left(k+1\right)^2 \label{2Dgd:eq:OepsilonFA}$ And the total operation results in

$M_{1n}= \sqrt{ \dfrac{3\,\left(k+1\right) -3 + k }{ 4\,k} } = 1 \label{2Dgd:eq:OsmallMaxDefA}$ Interesting to point out that as a consequence of this assumption the maximum shock angle, $$\theta$$ is a normal shock. However, taking the second term results in different value. Taking the second term in the explanation results in

$M_{1n}= \sqrt{ \dfrac{\sqrt{9\,\left(k+1\right)^2 +( 20\,k^2+8\,k-12) \,\epsilon} -3 + k + 2\,(1 + k) \epsilon }{ 4\,k} } \label{2Dgd:eq:OsmallMaxDefA1}$ Note this equation (46) produce an un realistic value and additional terms are required to obtained to produce a realistic value.

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