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Engineering LibreTexts 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 \(M_1 = \infty\).
    a_1 & = -1 - k \sin^2\delta \\
    a_2 & = \dfrac{\left( k + 1 \right)^ 2 \sin ^2 \delta }{ 4 } \\
    a_3 & = 0
    With these values the coefficients \(R\) and \(Q\) are
    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}
    Q = \dfrac{ ( 1 + k\, \sin^2\delta )^2 }{ 9 }

    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} \tag{31}
    The solution to this equation (31) for the weak angle is

    Weak Angle Solution

    \theta_{weak} = \tan^{-1}
    \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) \tag{32}

    Strong Angle Solution

    \theta_{strong} = \tan^{-1} \, \dfrac{

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    where these additional functions are
    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} \tag{34}
    f_2(M_1,\delta) = \dfrac{{ \left({M_1}^2 -1 \right)^3 - 9
    \left( 1 + \dfrac{k-1 }{ 2} {M_1}^{2}\right)
    \left( 1 + \dfrac{k-1 }{ 2} {M_1}^{2} + \dfrac{k+1 }{ 4} {M_1}^{4}
    \right) \tan^2\delta } }
    {{ f_1(M_1,\delta)^3} }
    \label{2Dgd:eq:Ofmtheta1a} \tag{35}
    Figure (??) exhibits typical results for oblique shock for two deflection angle of 5 and 25 degree. Generally, the strong shock is reduced as the increase of the Mach number while the weak shock is increase. The impossible shock for unsteady state is almost linear function of the upstream Mach number and almost not affected by the deflection angle.

    Fig. 12.6 The three different Mach numbers after the oblique

    The Procedure for Calculating The Maximum Deflection Point

    The maximum angle is obtained when D=0. When the right terms defined in (20)-(??), (??), and (??) are substituted into this equation and utilizing the trigonometrical identity \(\sin^2\delta + \cos^2\delta = 1\) and other trigonometrical identities results in Maximum Deflection Mach Number's equation in which is
    {M_1}^2 \,\left( k + 1 \right)\, ({M_{1n}}^2+1) =
    2\,(k\,{M_{1n}}^4 +2\,{M_{1n}}^2 - 1)
    \label{2Dgd:eq:Omenikoff} \tag{36}
    This equation and its twin equation can be obtained by an alternative procedure proposed by someone who suggested another way to approach this issue. It can be noticed that in equation (12), the deflection angle is a function of the Mach angle and the upstream Mach number, \(M_1\). Thus, one can conclude that the maximum Mach angle is only a function of the upstream Much number, \(M_1\). This can be shown mathematically by the argument that differentiating equation (12) and equating the results to zero creates relationship between the Mach number, \(M_1\) and the maximum Mach angle, \(\theta\). Since in that equation there appears only the heat ratio \(k\), and Mach number, \(M_1\), \(\theta_{max}\) is a function of only these parameters. The differentiation of the equation (12) yields
    \dfrac{d \tan \delta }{ d\theta} =
    {k {M_1}^4 \sin^4\theta +
    \left(2 - \dfrac{(k+1)}{ 2}{M_1}^2 \right) {M_1}^2\sin^2\theta
    -\left({ 1 + \dfrac{(k+1)}{ 2}{M_1}^2 } \right)
    \over k\, {M_1}^4 \sin^4\theta -
    \left[(k-1)+ \dfrac{(k+1)^2 {M_1}^2 }{ 4 } \right]
    {M_1}^2\sin^2\theta -1 }
    \label{2Dgd:eq:OsolD} \tag{37}
    Because tan is a monotonous function, the maximum appears when \(\theta\) has its maximum. The numerator of equation (37) is zero at different values of the denominator. Thus, it is sufficient to equate the numerator to zero to obtain the maximum. The nominator produces a quadratic equation for \(\sin^2\theta\) is applied here. Thus, the \(\sin^2\theta\) is
    \sin ^2 \theta_{max} =
    \dfrac{ -1 + i\dfrac{ k + 1 }{ 4}{M_1}^2+ \sqrt{(k+1)
    \left[ 1 + \dfrac{k-1 }{ 2} {M_1}^2 +
    \left( \dfrac{k+1 }{ 2} {M_1} \right)^4 \right]} }
    { k \, {M_1}^2}
    \label{2Dgd:eq:OthetaMax} \tag{38}
    Equation (38) should be referred to as the maximum's equation. It should be noted that both the Maximum Mach Deflection equation and the maximum's equation lead to the same conclusion that the maximum \(M_{1n}\) is only a function of upstream the Mach number and the heat ratio \(k\). It can be noticed that the Maximum Deflection Mach Number's equation is also a quadratic equation for \({M_{1n}}^2\). 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} \tag{39}
    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} \tag{40}
    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} =

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    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}
    {\sqrt{\dfrac{\sqrt{\epsilon (k) } + {\epsilon^2} +2\,\epsilon -
    3 + k\, \epsilon^2+2\,k\epsilon+k }{ 4\,k} } } \tag{42}

    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} \tag{43}
    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} \tag{44}
    And the total operation results in
    M_{1n}= \sqrt{ \dfrac{3\,\left(k+1\right) -3 + k }{ 4\,k} } = 1
    \label{2Dgd:eq:OsmallMaxDefA} \tag{45}
    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} \tag{46}
    Note this equation (46) produce an un realistic value and additional terms are required to obtained to produce a realistic value.


    • Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.