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

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    847
  • [ "article:topic" ]

    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 \(M_1\)

    \[
        \label {2Dgd:eq:Odelta-theta}
        \cot \delta = \tan \left(\theta\right) \left[
                \dfrac{(k + 1)\, {M_1}^2 }{ 2\, ( {M_1}^2\,  \sin^2 \theta - 1)} - 1
            \right]  \tag{51}
    \]

    or
    \[
        \tan \delta = {2\cot\theta ({M_1}^2 \sin^2 \theta -1 ) \over  
            2 + {M_1}^2 (k + 1 - 2 \sin^2 \theta )}  
        \label{2Dgd:eq:Odelta-thetaA}  \tag{52}
    \]
    The pressure ratio can be expressed as

    Pressure Ratio

    \[
    \label {2Dgd:eq:OPR}  
        \dfrac{P_ 2 }{ P_1} = \dfrac{ 2 \,k\, {M_1}^2 \sin ^2 \theta - (k -1) }{ k+1}  \tag{53}
    \]

    The density ratio can be expressed as

    Density Ratio

    \[
        \label {2Dgd:eq:ORR}
        \dfrac{\rho_2 }{ \rho_1 } = \dfrac{ {U_1}_n }{ {U_2}_n}     
        = \dfrac{ (k +1)\, {M_1}^2\, \sin ^2 \theta }
            { (k -1) \, {M_1}^2\, \sin ^2 \theta + 2}    \tag{54}
    \]

    The temperature ratio expressed as

    Temperature Ratio

    \[
        \label {2Dgd:eq:OTR}
        \dfrac{ T_2 }{ T_1} = \dfrac{ {c_2}^2 }{ {c_1}^2} =  
        \dfrac{ \left( 2\,k\, {M_1}^2 \sin ^2 \theta - ( k-1) \right)   
            \left( (k-1) {M_1}^2 \sin ^2 \theta + 2 \right) }
            { (k+1)\, {M_1}^2\, \sin ^2 \theta  }  \tag{55}
    \]

    The Mach number after the shock is

    Exit Mach Number

    \[
        \label{2Dgd:eq:OM2_0}
        {M_2}^2 \sin (\theta -\delta) =  
        { (k -1) {M_1}^2 \sin ^2 \theta +2  \over  
            2 \,k\,  {M_1}^2 \sin ^2 \theta - (k-1) }  \tag{56}
    \]

    or explicitly
    \[
        {M_2}^2 = {(k+1)^2 {M_1}^4 \sin ^2 \theta  -
            4\,({M_1}^2 \sin ^2 \theta  -1) (k {M_1}^2 \sin ^2 \theta +1)
        \over
            \left( 2\,k\, {M_1}^2 \sin ^2 \theta - (k-1) \right)
            \left(  (k-1)\, {M_1}^2 \sin ^2 \theta +2 \right)
        }   
        \label{2Dgd:eq:OM2}  \tag{57}
    \]
    The ratio of the total pressure can be expressed as

    Stagnation Pressure Ratio

    \[
        \label {2Dgd:eq:OP0R}
        {P_{0_2} \over P_{0_1}} = \left[
            (k+1) {M_1}^2 \sin ^2 \theta \over  
            (k-1) {M_1}^2 \sin ^2 \theta +2 \right]^{k \over k -1}     
            \left[ k+1 \over 2 k {M_1}^2 \sin ^2 \theta - (k-1) \right]
                    ^{1 \over k-1}  \tag{58}
    \]

    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

    • 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.