# What is the Mach number downstream of the expansion corner?

Consider the inviscid,adiabatic flow of air at free stream conditions $$M_{1}=2,P_{1}=1\;atm$$ and $$T_{1}=288\;K$$ around a sharp expansion corner $$\left ( \theta =20^{\circ} \right )$$ as shown below.The Prandtl-Meyer function $$\nu$$,is given as a function of Mach number M , as

$\nu (M)=\sqrt{\frac{\gamma +1}{\gamma -1}}tan^{-1}\sqrt{\frac{\gamma -1}{\gamma +1}\left ( M^{2} -1\right )}-tan^{-1}\sqrt{M^{2}-1}$

Assume air to be calorically perfect with $$\gamma =1.4$$ .What is the Mach number $$M_{2}$$, downstream of the expansion corner ?

Figure

Asked on 14th October 2019 in

Prandtl-Meyer expansion theory tells that $\theta =\nu \left ( M_{2} \right )-\nu \left ( M_{1} \right )$

Here

$$\nu \left ( M_{1} \right )=\sqrt{\frac{2.4}{0.4}}tan^{-1}\sqrt{\frac{0.4}{2.4}\times 3}-tan^{-1}\sqrt{3} \\=26.38^{\circ}$$

$$\nu \left ( M_{2} \right )=\theta +\nu \left ( M_{1} \right ) \\=20^{\circ}+26.38^{\circ} \\=46.35^{\circ}$$

On putting the value of $$\nu (M_{2})$$ in the Prandtl-Meyer function we get

$$46.35^{\circ}=\sqrt{\frac{2.4}{0.4}}tan^{-1}\sqrt{\frac{0.4}{2.4}\left ( M_{2}^{2}-1 \right )}-tan^{-1}\sqrt{M_{2}^{2}+1}$$

$$\Rightarrow M_{2}=2.83$$ approximately.