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flow with M = 3 comes straight on in the x-direction towards the wedge. We know theta from our geometry of the wedge. From this we can calculate the normal component of our free stream Mach number.
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\LARGE \begin{equation}\nonumber M_{1N} = M_1sin(\beta) \end{equation} \\ |
Where the shock angle Beta comes from the theta-beta-M chart.
Now we can relate the normal Mach numbers to each other through the normal shock relations
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\LARGE \begin{equation}\nonumber M_{2N}^2 = M_{1N}^2(\frac{(\gamma -1)M_{1N}+2}{2\gamma M_{1N}-(\gamma -1)}) \end{equation} \\ \\ \begin{equation}\nonumber \frac{p2}{p1} = \frac{2\gamma M_{1N}^2 - (\gamma - 1)}{\gamma + 1} \end{equation} \\ \\ \begin{equation}\nonumber \frac{T2}{T1} = \frac{(2\gamma M_{1N}^2 - (\gamma - 1))((\gamma -1)M_{1N}^2 +2)}{(\gamma +1)^2 M_{1N}^2} \end{equation} \\ |
From the equations above it is quite easy to relate upstream quantities.
We also expect that the flow downstream of the shock will still be supersonic as the flow experiences only a weak oblique shock, evident from looking at the theta-beta-M chart. This also becomes clear in the hand calculations.
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