Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 314 / 2 / a

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Solution

In the shock frame, integrate mass conservation, momentum conservation, and total-energy conservation across a thin interval containing the stationary normal shock wave. No mass, momentum, or energy is stored in the vanishingly thin interval, so each flux has the same value on both sides. For a perfect gas, the specific enthalpy is

h = γ p / [(γ - 1) ρ]

. The resulting Rankine-Hugoniot conditions for a perfect gas are

ρ_{1} u_{1} p_{1} + ρ_{1} u_{1}^{2} u_{1} (\frac{ρ _{1} u _{1}^{2}}{2} + \frac{γ p _{1}}{γ - 1}) = ρ_{2} u_{2}, = p_{2} + ρ_{2} u_{2}^{2}, = u_{2} (\frac{ρ _{2} u _{2}^{2}}{2} + \frac{γ p _{2}}{γ - 1}) .

The momentum flux includes both transported momentum and pressure force. The energy flux includes kinetic energy and enthalpy, the latter accounting for internal energy and the work needed to push gas through the interval. Entropy production can occur inside the normal shock wave even though total energy is conserved.

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