A simple analytical solution to the Sedov-Taylor similarity equations can be found for the particular case γ = 7. This is a fair approximation to the behavior of water under explosive conditions, as it will be almost incompressible.

(a) Make the ansatz (whose self-consistency we will check later) that the velocity in the post-shock flow varies linearly with radius from the origin to the shock: ṽ(ξ) = ξ .Use Eq. (17.45) to transform the equation of continuity into an ordinary differential equation and hence solve for the density function ρ̃(ξ).

(b) Next use the equation of motion to discover that P̃(ξ) = ξ³.

(c) Verify that your solutions for the functions P̃, ρ̃, and ṽ satisfy the remaining entropy equation, thereby vindicating the original ansatz.

(d) Substitute your results from parts (a) and (b) into Eq. (17.48) to show that

(e) An explosive charge weighing 100 kg with an energy release of 10⁸ J kg⁻¹ is detonated underwater. For what range of shock radii do you expect the Sedov-Taylor similarity solution to be valid?

Equations 17.45 and 17.48.

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E = 2π RS PO R5 Po 225t²