Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 314 / 2

For one-dimensional compressible flow, define the total energy densityThe conservative mass, momentum, and energy equations areA monatomic nonrelativistic perfect gas has $\gamma =5/3$, an ultrarelativistic gas or radiation-dominated fluid has $\gamma =4/3$, and a rotationally active diatomic gas has $\gamma =7/5$.

Integrating the three conservation laws across a stationary normal shock wave gives the Rankine-Hugoniot conditions for a perfect gasEliminate $u_{x2}$ with the conserved mass flux and write $v_s^2=\gamma p/\rho$ and $M_x=u_x/v_s$. Solving the remaining two algebraic equations givesFor a compressive shock, $M_{x1}>1$, the density and pressure increase, and the downstream normal flow is subsonic in the shock frame.

For an oblique shock, boost parallel to the front by the upstream tangential speed. The transformed upstream flow is normal, so part b applies. Inviscid momentum balance has no tangential stress and therefore makes the tangential velocity continuous. Transforming back giveswhile all normal velocity, density, and pressure relations from part b remain unchanged with $M_x$ the normal Mach number.

The pressure jump relation with $p_2/p_1=1+ϵ$ givesIf ${\beta }_1$ is the angle between the upstream velocity and the shock front, then $M_{x1}=M_1\sin {\beta }_1$. To leading order,the Mach angle. Expansion of the compression ratio givesBecause $u_y$ is continuous and $\tan \beta =u_x/u_y$,Writing $\theta ={\beta }_1-{\beta }_2$ and linearizing the tangent about $\sin {\beta }_1=1/M_1$ gives the weak-oblique-shock deflectionThe normal component decreases while the tangential component is unchanged, so ${\beta }_2<{\beta }_1$: the flow turns toward the shock front.