Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 314 / 2

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

The dimensions of the injected power and ambient mass density are $[\dot{E}]=M{L}^2{T}^{-3}$ and $[{\rho }_1]=M{L}^{-3}$. Requiring ${\dot{E}}^a{\rho }_1^b{t}^c$ to have the dimensions of length givesThus dimensional analysis gives the adiabatic superbubble similarity solutionThe dimensionless coefficient $\alpha$ can depend on $\gamma$ and the specified inner structure, but dimensions alone cannot determine it. Constant power differs from a fixed-energy explosion because the energy available at time $t$ is proportional to $t$. This scaling assumes that ambient pressure, gravity, radiative losses and the finite injection radius introduce no relevant competing scale.

The original PDF supplies the intended profiles $\rho ={\rho }_{1}f(\eta )$, $p={\rho }_1{\dot{R}}^{2}g(\eta )$ and $u=\dot{R}h(\eta )$; the TeX has corrupted the first two. At fixed $r$, ${\partial }_t\eta =-\eta \dot{R}/R$, and at fixed $t$, ${\partial }_r\eta =1/R$. The spherical continuity equation is thereforeSince $\dot{R}/R=3/(5t)$ is nonzero, the continuity equation for a superbubble similarity solution isThe ambient gas is at rest. In the locally stationary shock frame, its velocity is $-\dot{R}$; the downstream velocity is $u-\dot{R}$. Using the shock compression ratio from part (a) converts the downstream speed back to the laboratory frame and gives the inner shock boundary conditionsImmediately outside the shock, the corresponding unperturbed profiles are $f=1$, $g=0$ and $h=0$, with $g=0$ understood in the negligible-ambient-pressure limit.

The inner boundary must supply the injected power. If the source is idealized as a point and the profiles extend all the way inward, central energy input in a similarity solution requiresor equivalently ${\lim }_{\eta \downarrow 0}{\eta }^{2}h(f{h}^2/2+\gamma g/(\gamma -1))=125/(108\pi {\alpha }^5)$. A finite source region or an inner wind/contact region supplies a different matching description. One cannot impose a regular zero-flux centre on a source-free interior and still inject nonzero power.

The conservation of energy to use here is the total kinetic energy plus internal energy inside the shock, not just the internal energy of a uniform-pressure core. With no radiative loss and negligible initial ambient energy,Substitute the adiabatic superbubble similarity solution and define the finite positive dimensionless integralThen $\dot{E}t=4\pi {\rho }_1{\dot{R}}^2{R}^{3}I$. Since $\dot{R}=3R/(5t)$ and $R^5={\alpha }^5(\dot{E}/{\rho }_1)t^3$,This is the energy normalization of a continuously driven spherical shock. It fixes $\alpha$ once the similarity profiles are determined. If the profiles describe only the shocked ambient layer, the energy of the interior hot bubble must also be included before using this normalization. A radiatively cooled shell has a different energy budget; its familiar numerical coefficient should not be substituted into this loss-free model.