An oblique shock stands at angle to an upstream stream of Mach number

PHYSICS · Subjective · Fermi University Physics · Fluids & Continuum Physics · Waves, acoustics and compressible flow · Waves, acoustics and compressible flow

Signal Speeds in Fluids, Ch Oblique shocks, expansion fans and blast waves

Problem

An oblique shock stands at angle β\beta to an upstream stream of Mach number M1M_1, and deflects the stream through an angle θ\theta towards the shock. Only the velocity component normal to the shock front is changed by the jump; the tangential component is unchanged, and the normal component upstream and downstream is related through Chapter 8's normal-shock density ratio ρ2ρ1=(γ+1)M1n2(γ−1)M1n2+2,M1n=M1sin⁡β.\frac{\rho_2}{\rho_1} = \frac{(\gamma+1)M_{1{\rm n}}^2}{(\gamma-1)M_{1{\rm n}}^2+2}, \qquad M_{1{\rm n}} = M_1\sin\beta. (a) Draw the velocity triangle upstream and downstream of the shock (normal component unu_{\rm n}, tangential component utu_{\rm t}, unchanged across the shock) and write tan⁡β\tan\beta and tan⁡(β−θ)\tan(\beta-\theta) in terms of u1nu_{1{\rm n}}, u2nu_{2{\rm n}} and utu_{\rm t}. (b) Use mass conservation across the shock to write tan⁡(β−θ)/tan⁡β\tan(\beta-\theta)/\tan\beta in terms of ρ1/ρ2\rho_1/\rho_2, and hence in terms of M1n2M_{1{\rm n}}^2. (c) Expand tan⁡(β−θ)\tan(\beta-\theta) with the compound-angle formula and solve for tan⁡θ\tan\theta, then use M1n=M1sin⁡βM_{1{\rm n}}=M_1\sin\beta and the double-angle identity cos⁡2β=cos⁡2β−sin⁡2β\cos2\beta = \cos^2\beta-\sin^2\beta to reach tan⁡θ=2cot⁡β (M12sin⁡2β−1)M12(γ+cos⁡2β)+2.\tan\theta = \frac{2\cot\beta\,(M_1^2\sin^2\beta-1)}{M_1^2(\gamma+\cos2\beta)+2}.

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