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

- Page ID
- 847

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}

\]

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}

\]

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.