Someone argues as follows

PHYSICS · Multi-part · Fermi University Physics · Fluids & Continuum Physics · Thermal, Statistical & Complex Systems · The laws of thermodynamics

Signal Speeds in Fluids, Ch Steepening and the normal shock wave

Problem

Someone argues as follows. "The normal shock relations for air (γ=1.4\gamma = 1.4) give, for M1=2M_1 = 2: M2=0.577M_2 = 0.577, p2/p1=4.50p_2/p_1 = 4.50, ρ2/ρ1=2.67\rho_2/\rho_1 = 2.67, T2/T1=1.69T_2/T_1 = 1.69. These numbers satisfy mass, momentum and energy conservation across the jump. But those three conservation statements do not care which side is labelled 1 and which is labelled 2: if I simply relabel the states, swapping upstream and downstream, I get an equally valid discontinuity with upstream Mach number M1′=0.577M_1' = 0.577 and downstream M2′=2M_2' = 2, and ratios p2′/p1′=1/4.50=0.222p_2'/p_1' = 1/4.50 = 0.222, ρ2′/ρ1′=1/2.67=0.375\rho_2'/\rho_1' = 1/2.67 = 0.375, T2′/T1′=1/1.69=0.592T_2'/T_1' = 1/1.69 = 0.592. Since mass, momentum and energy are conserved in both directions, this reversed jump, which expands and accelerates subsonic air to M=2M = 2, must be just as physically admissible as the original compressive one." ((a)) Compute the entropy change Δs/R\Delta s/R across the relabelled (reversed) jump, using Δs/R=γγ−1ln⁡(T2′/T1′)−ln⁡(p2′/p1′)\Delta s/R = \dfrac{\gamma}{\gamma - 1}\ln(T_2'/T_1') - \ln(p_2'/p_1'). ((b)) State exactly where the argument fails. Does any of the three conservation statements distinguish the two directions, and if not, which further law does?

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