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

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.