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Shock stand-off distance for blunt body

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Old   August 13, 2003, 10:59
Default Shock stand-off distance for blunt body
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Praveen
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There is an empirical formula for calculating shock stand-off distance for a 2-D cylinder and 3-D hemisphere as a function of Mach number. Does anybody have these formulae ? I think these are given in Anderson's Modern Compressible Flow, but I dont have access to that book. I would be grateful if somebody can give me the formulae. Thanks
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Old   August 13, 2003, 22:29
Default Re: Shock stand-off distance for blunt body
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Michael
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Hypersonic and High Temperature Gas Dynamics (Anderson, Page 189:

5.4 Correlations for Hypersonic Shock-Wave Shapes

Here's a summary:

Shock shape is given by the equation:

x = R + delta - Rc*cot^2[beta*(sqrt(1+(y^2*tan^2(beta))/Rc^2)-1 )]

where:

delta/R =

0.143*exp(3.24/Minfinity^2) for sphere-cone 0.386*exp[4.67/Minfinity^2) for cylinder-wedge

and

Rc/R =

1.143*exp(0.54/(Minfinity-1)^1.2) for sphere-cone 1.386*exp(1.8/(Minfinity-1)^0.75) for cylinder-wedge

For cartesion cooridinates (x,y), where delta is the stand off distance and R is the sphere / cylinder radius and beta is the limiting angle to which the curved bow shock is asymtopic. Rc is the radius of curvature at the vertex of the hyperbola. The curves are for a sphere-cone (axisymmetric) and cylinder-wedge (planar) combination. The cone/wedge angle is 14.03624 degrees.
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