Upstream Mach Number, M1, and Shock Angle, θ

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

$$\cot \delta = \tan \theta \left[ {(k + 1) {M_1}^2 \over 2 ( {M_1}^2 \sin^2 \theta -1)} - 1 \right]$$ (13.51)

or

$$\tan \delta = {2\cot\theta ({M_1}^2 \sin^2 \theta -1 ) \over 2 + {M_1}^2 (k + 1 - 2 \sin^2 \theta )}$$ (13.52)

The pressure ratio can be expressed as

$${P_ 2 \over P_1} = { 2 k {M_1}^2 \sin ^2 \theta - (k -1) \over k+1}$$ (13.53)

The density ratio can be expressed as

$${\rho_2 \over \rho_1 } = { {U_1}_n \over {U_2}_n} = { (k +1) {M_1}^2 \sin ^2 \theta \over (k -1) {M_1}^2 \sin ^2 \theta + 2}$$ (13.54)

The temperature ratio expressed as

$${ T_2 \over T_1} = { {c_2}^2 \over {c_1}^2} = { \left( 2k {M_1}^2... ...( (k-1) {M_1}^2 \sin ^2 \theta + 2 \right) \over (k+1) {M_1}^2 \sin ^2 \theta }$$ (13.55)

The Mach number after the shock is

$${M_2}^2 \sin (\theta -\delta) = { (k -1) {M_1}^2 \sin ^2 \theta +2 \over 2 k {M_1}^2 \sin ^2 \theta - (k-1) }$$ (13.56)

or explicitly

$${M_2}^2 = {(k+1)^2 {M_1}^4 \sin ^2 \theta - 4({M_1}^2 \sin ^2 \th... ...\sin ^2 \theta - (k-1) \right) \left( (k-1) {M_1}^2 \sin ^2 \theta +2 \right) }$$ (13.57)

The ratio of the total pressure can be expressed as

$${P_{0_2} \over P_{0_1}} = \left[ (k+1) {M_1}^2 \sin ^2 \theta \ov... ... -1} \left[ k+1 \over 2 k {M_1}^2 \sin ^2 \theta - (k-1) \right] ^{1 \over k-1}$$ (13.58)

Even though the solution for these variables, $M_1$ and $\theta$ , is unique, the possible range deflection angle, $\delta$ , is limited. Examining equation (13.51) 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 13.9). The range of given $\theta$ , upstream Mach number $M_1$ , is limited between $\infty$ and $\sqrt{1 / \sin^{2}\theta}$ .

Figure 13.9: The possible range of solutions