Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 70 / 1

Write $v=u_x$ and $D={\partial }_t+v{\partial }_x$. The Euler equations and conservation of mass give $D\rho =-\rho v_x$ and $\rho Du=-p_x{e}_x$. The internal-energy density of the ideal gas is $e=p/(\gamma -1)$. The adiabatic pressure equation therefore givesMultiply the momentum equation by $u$ and use continuity to obtain the kinetic-energy balanceAdding the two equations moves ${\partial }_x(pv)$ into the flux. Thus the entire system is ${\partial }_{t}Q+{\partial }_{x}F=0$, whereThis derives the conservative energy flux of a polytropic ideal gas. In particular one-dimensional spatial dependence does not remove the transverse velocities from the total energy. The energy flux can equally be written $\rho v(u^2/2+w)$ with specific enthalpy $w=\gamma p/[(\gamma -1)\rho ]$.

Integrate any conservation equation from part (a) through a vanishingly thin control volume moving with shock speed $s$. Its singular terms require $[F]=s[Q]$. The stationary shock has $s=0$, so all five fluxes are continuous. Substitution gives the Rankine-Hugoniot conditions for a perfect gas:Let the common mass flux be $j={\rho }_1{v}_1={\rho }_2{v}_2>0$. Dividing the transverse conditions by $j$ proves $u_{y2}=u_{y1}$ and $u_{z2}=u_{z1}$. Dividing the energy condition by $j$ shows that $u^2/2+w$ is unchanged. Thus tangential kinetic energies cancel when solving for the normal shock compression. No global equality of the entropy parameter $p/{\rho }^{\gamma }$ across the shock has been assumed: physical dissipative shocks increase that parameter.

Set $a=v_2/v_1$. Mass and normal momentum conservation implyAfter cancelling the unchanged tangential kinetic energies, the energy condition is $v_1^2/2+\gamma p_1/[(\gamma -1){\rho }_1]=a^2{v}_1^2/2+\gamma a{p}_2/[(\gamma -1){\rho }_1]$. With $v_{s1}^2=\gamma p_1/{\rho }_1$ and normal Mach number $\mathcal{M}=v_1/v_{s1}$, rearrangement factors it asThe first root is the no-shock solution. The nontrivial perfect-gas shock jump conditions therefore giveIts density and pressure ratios arePhysical admissibility selects a compressive, entropy-increasing shock. This can be verified directly: with $X={\mathcal{M}}^2$, define $K_i=p_i/{\rho }_i^{\gamma }$. The derivative of $\ln (K_2/K_1)$ after inserting these ratios iswhere pressures are positive, and $K_2/K_1=1$ at $X=1$. Thus the entropy increases precisely on the branch $X>1$; the positive-pressure expansion branch $X<1$ decreases entropy. Since the normal speed is positive,The endpoint one is the zero-strength limit, not a finite shock. The downstream normal Mach number obeys ${\mathcal{M}}_2^2=[(\gamma -1){\mathcal{M}}^2+2]/[2\gamma {\mathcal{M}}^2-(\gamma -1)]<1$.

Let $U=|u_1|$ and $q=(\gamma -1)/(\gamma +1)$, the strong-shock limit of the normal velocity ratio. In the plane containing the upstream direction and the shock normal, take $u_1=U{e}_X$ and $n=\cos \beta e_X+\sin \beta e_Y$. The normal velocity is multiplied by $q$ while the tangential velocity is unchanged, henceEliminating $\beta$ gives the strong-shock polar for a perfect gas:Equivalently this is the circleBoth signs of the perpendicular component are possible by changing shock orientation. The endpoints $u_{X2}=qU,U$ lie on the upstream axis. An original sketch for $\gamma =5/3$ illustrates the locus and a tangent ray:

The deflection is the angle of the ray from the origin to the downstream velocity, rather than the shock-normal angle $\beta$. Its maximum occurs where this ray is tangent to the circle. The circle's center is at distance $\gamma U/(\gamma +1)$ and its radius is $U/(\gamma +1)$, so the tangent right triangle provesThis is the maximum deflection through a strong perfect-gas shock; for $\gamma =5/3$ it is about $36.87$ degrees.