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12.2.2.3: Upstream Mach Number, \(M_1\), and Shock Angle, \(\theta\)

  • Page ID
    847

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    The solution for upstream Mach number, \(M_1\), and shock angle, θ, are far much simpler and a unique solution exists. The deflection angle can be expressed as a function of these variables as

    \(\delta\) For \(\theta\) and \(M_1\)

    \[ \label {2Dgd:eq:Odelta-theta}
    \cot \delta = \tan \left(\theta\right) \left[
    \dfrac{(k + 1)\, {M_1}^2 }{ 2\, ( {M_1}^2\, \sin^2 \theta - 1)} - 1
    \right]
    \]

    or

    \[ \tan \delta = {2\cot\theta ({M_1}^2 \sin^2 \theta -1 ) \over
    2 + {M_1}^2 (k + 1 - 2 \sin^2 \theta )}
    \label{2Dgd:eq:Odelta-thetaA}
    \]

    The pressure ratio can be expressed as

    Pressure Ratio

    \[ \label {2Dgd:eq:OPR}
    \dfrac{P_ 2 }{ P_1} = \dfrac{ 2 \,k\, {M_1}^2 \sin ^2 \theta - (k -1) }{ k+1}
    \]

    The density ratio can be expressed as

    Density Ratio

    \[ \label {2Dgd:eq:ORR}
    \dfrac{\rho_2 }{ \rho_1 } = \dfrac{ {U_1}_n }{ {U_2}_n}
    = \dfrac{ (k +1)\, {M_1}^2\, \sin ^2 \theta }
    { (k -1) \, {M_1}^2\, \sin ^2 \theta + 2}
    \]

    The temperature ratio expressed as

    Temperature Ratio

    \[ \label {2Dgd:eq:OTR}
    \dfrac{ T_2 }{ T_1} = \dfrac{ {c_2}^2 }{ {c_1}^2} =
    \dfrac{ \left( 2\,k\, {M_1}^2 \sin ^2 \theta - ( k-1) \right)
    \left( (k-1) {M_1}^2 \sin ^2 \theta + 2 \right) }
    { (k+1)\, {M_1}^2\, \sin ^2 \theta }
    \]

    The Mach number after the shock is

    Exit Mach Number

    \[ \label{2Dgd:eq:OM2_0}
    {M_2}^2 \sin (\theta -\delta) =
    { (k -1) {M_1}^2 \sin ^2 \theta +2 \over
    2 \,k\, {M_1}^2 \sin ^2 \theta - (k-1) }
    \]

    or explicitly

    \[ {M_2}^2 = {(k+1)^2 {M_1}^4 \sin ^2 \theta -
    4\,({M_1}^2 \sin ^2 \theta -1) (k {M_1}^2 \sin ^2 \theta +1)
    \over
    \left( 2\,k\, {M_1}^2 \sin ^2 \theta - (k-1) \right)
    \left( (k-1)\, {M_1}^2 \sin ^2 \theta +2 \right)
    }
    \label{2Dgd:eq:OM2}
    \]

    The ratio of the total pressure can be expressed as

    Stagnation Pressure Ratio

    \[ \label {2Dgd:eq:OP0R}
    {P_{0_2} \over P_{0_1}} = \left[
    (k+1) {M_1}^2 \sin ^2 \theta \over
    (k-1) {M_1}^2 \sin ^2 \theta +2 \right]^{k \over k -1}
    \left[ k+1 \over 2 k {M_1}^2 \sin ^2 \theta - (k-1) \right]
    ^{1 \over k-1}
    \]

    Even though the solution for these variables, \(M_1\) and \(\theta\), is unique, the possible range deflection angle, \(\delta\), is limited. Examining equation (51) shows that the shock angle, \(\theta\), has to be in the range of \(\sin^{-1} (1/M_1) \geq \theta \geq (\pi/2)\) (see Figure Fig. 12.8). The range of given \(\theta\), upstream Mach number \(M_1\), is limited between \(\infty\) and \(\sqrt{1 / \sin^{2}\theta}\).

    Fig. 12.8 The possible range of solutions for different parameters for given upstream Mach numbers.

    Contributors and Attributions

    • 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.


    This page titled 12.2.2.3: Upstream Mach Number, \(M_1\), and Shock Angle, \(\theta\) is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by via source content that was edited to the style and standards of the LibreTexts platform.


    This page titled 12.2.2.3: Upstream Mach Number, \(M_1\), and Shock Angle, \(\theta\) is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Genick Bar-Meir via source content that was edited to the style and standards of the LibreTexts platform.